In a class there are $12$ boys and $10$ girls. The teacher wants to select either a boy or a girl to represent the class in a competition. In how many ways can this be done?
(A) $120$
(B) $22$
(C) $2$
(D) $20$
View Step-by-Step Mathematical Solution
Answer: (B)
The teacher has to select **either** a boy **or** a girl.
By the **Fundamental Principle of Addition**:
$$\text{Total ways} = 12 + 10 = 22$$
Hence, option **(ii) / (B)** is correct.
A person predicts the outcome of $4$ successive football matches. Each match can end in one of three ways: Win, Draw, or Loss. The total number of different prediction sequences for the $4$ matches is:
(A) $81$
(B) $64$
(C) $9$
(D) $63$
View Step-by-Step Mathematical Solution
Answer: (A)
For each match, there are $3$ possible predicted outcomes (Win, Draw, Loss).
Since the predictions for each of the $4$ successive matches are independent sequential events, by the **Multiplication Principle of Counting**:
$$\text{Total prediction sequences} = 3 \times 3 \times 3 \times 3 = 3^4 = 81$$
Hence, option **(i) / (A)** is correct.
Number of ways in which $15$ different children can sit in a merry-go-round relative to one another is:
(A) $\frac{1}{2} (14!)$
(B) $\frac{1}{2} (15!)$
(C) $14!$
(D) $2 \times 14!$
View Step-by-Step Mathematical Solution
Answer: (C)
The number of circular permutations of $n$ distinct objects seated around a circle when clockwise and counter-clockwise arrangements are distinct is:
$$P_c = (n - 1)!$$
For $n = 15$ children:
$$\text{Number of ways} = (15 - 1)! = 14!$$
Hence, option **(iii) / (C)** is correct.
Number of diagonals of a convex hexagon are:
(A) $11$
(B) $9$
(C) $6$
(D) $15$
View Step-by-Step Mathematical Solution
Answer: (B)
The number of diagonals in an $n$-sided convex polygon is:
$$D = \,^nC_2 - n = \frac{n(n-3)}{2}$$
For a hexagon ($n = 6$):
$$D = \frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9$$
Hence, option **(ii) / (B)** is correct.
Number of divisors of $10,000,000$ are:
(A) $12$
(B) $49$
(C) $8$
(D) $64$
View Step-by-Step Mathematical Solution
Answer: (D)
Express $10,000,000 = 10^7$ in prime factorization:
$$10^7 = 2^7 \times 5^7$$
Total number of divisors is given by $(a_1 + 1)(a_2 + 1)$:
$$\text{Total divisors} = (7 + 1)(7 + 1) = 8 \times 8 = 64$$
Hence, option **(iv) / (D)** is correct.
A donuts shop offers $20$ kinds of donuts. The shop has at least a dozen donuts of each kind. If a person enters the shop, in how many ways can he select a dozen ($12$) donuts?
(A) $^{31}C_{12}$ ways
(B) $^{30}C_{12}$ ways
(C) $^{32}C_{12}$ ways
(D) $240$ ways
View Step-by-Step Mathematical Solution
Answer: (A)
Selecting $r = 12$ items from $n = 20$ varieties with repetition allowed:
$$\text{Ways} = \,^{n + r - 1}C_r = \,^{20 + 12 - 1}C_{12} = \,^{31}C_{12}$$
Hence, option **(i) / (A)** is correct.
The number of permutations of $n$ different things taken $r$ at a time in which $m$ particular things are placed in $m$ given places in definite order is:
(A) $^{n-m}P_{r-m} \times m!$
(B) $(n - m + 1)!$
(C) $^{n-m}P_{r-m}$
(D) $^nP_r - m!$
View Step-by-Step Mathematical Solution
Answer: (C)
Since the $m$ particular things occupy $m$ fixed specific places in a **definite order**, there is only $1$ way to place them.
We then choose and arrange $(r - m)$ items from the remaining $(n - m)$ items:
$$\text{Total ways} = \,^{n-m}P_{r-m}$$
Hence, option **(iii) / (C)** is correct.
If $^{18}C_r = \,^{18}C_{r+2}$, find $^rC_5$:
(A) $56$
(B) $48$
(C) $54$
(D) None of these
View Step-by-Step Mathematical Solution
Answer: (A)
Using $^{18}C_x = \,^{18}C_y \implies x + y = 18$:
$$r + (r + 2) = 18 \implies 2r = 16 \implies r = 8$$
Now, $^rC_5 = \,^8C_5 = \,^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. Option **(i) / (A)** is correct.
If $^{15}C_{3r} = \,^{15}C_{r+3}$, then $r$ is equal to:
(A) $5$
(B) $4$
(C) $3$
(D) $2$
View Step-by-Step Mathematical Solution
Answer: (C)
Using $^{15}C_x = \,^{15}C_y \implies 3r + (r + 3) = 15 \implies 4r = 12 \implies r = 3$. Option **(iii) / (C)** is correct.
Total number of words formed by $2$ vowels and $3$ consonants taken from $4$ vowels and $5$ consonants is equal to:
(A) $60$
(B) $120$
(C) $7200$
(D) None of these
View Step-by-Step Mathematical Solution
Answer: (C)
1. Selection: $^4C_2 \times \,^5C_3 = 6 \times 10 = 60$.
2. Arrangement of 5 letters: $5! = 120$.
$$\text{Total words} = 60 \times 120 = 7200$$
Option **(iii) / (C)** is correct.
There are $12$ points in a plane. The number of straight lines joining any two of them when $3$ of them are collinear is:
(A) $62$
(B) $63$
(C) $64$
(D) $65$
View Step-by-Step Mathematical Solution
Answer: (C)
Lines $= \,^nC_2 - \,^pC_2 + 1 = \,^{12}C_2 - \,^3C_2 + 1 = 66 - 3 + 1 = 64$. Option **(iii) / (C)** is correct.
There are $5$ vacant seats in a row. In how many ways can $3$ men sit?
(A) $60$
(B) $15$
(C) $243$
(D) $125$
View Step-by-Step Mathematical Solution
Answer: (A)
$$\text{Ways} = \,^5P_3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60$$
Option **(i) / (A)** is correct.
Find the total number of ways of answering $6$ multiple choice questions, if each question has $4$ choices.
(A) $6^4$
(B) $6^6$
(C) $24$
(D) $4^6$
View Step-by-Step Mathematical Solution
Answer: (D)
Each of the $6$ questions has $4$ possible options.
By the Multiplication Principle of Counting:
$$\text{Total ways} = 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^6 = 4096$$
Hence, option **(iv) / (D)** is correct.
Compute $\frac{n!}{(n-r)!}$ when:
(i) $n = 8, r = 2$
(ii) $n = 12, r = 3$
View Detailed Derivation & Solution
Key: (i) 56, (ii) 1320
**Part (i):** For $n = 8, r = 2$:
$$\frac{8!}{(8-2)!} = \frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56$$
**Part (ii):** For $n = 12, r = 3$:
$$\frac{12!}{(12-3)!} = \frac{12!}{9!} = 12 \times 11 \times 10 = 1320$$
Find $r$ if $^9P_r = 3024$.
View Detailed Derivation & Solution
Key: $r = 4$
We know $^9P_r = 9 \times 8 \times 7 \times \dots$ ($r$ factors).
Factoring 3024 successively starting with 9:
- $3024 \div 9 = 336$
- $336 \div 8 = 42$
- $42 \div 7 = 6$
- $6 \div 6 = 1$
Thus:
$$3024 = 9 \times 8 \times 7 \times 6 = \,^9P_4$$
Therefore, $r = 4$.
Find $r$ if $^5P_r = 2 \cdot \,^6P_{r-1}$.
View Detailed Derivation & Solution
Key: $r = 3$
Using the definition $^nP_k = \frac{n!}{(n-k)!}$:
$$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(6-(r-1))!} = 2 \cdot \frac{6 \cdot 5!}{(7-r)!}$$
Cancelling $5!$ from both sides:
$$\frac{1}{(5-r)!} = \frac{12}{(7-r)(6-r)(5-r)!}$$
$$(7-r)(6-r) = 12$$
$$r^2 - 13r + 42 = 12 \implies r^2 - 13r + 30 = 0$$
$$(r - 10)(r - 3) = 0$$
Since $r \le 5$, we discard $r = 10$.
Hence, $r = 3$.
Convert the following products into factorials:
(i) $6 \cdot 7 \cdot 8 \cdot 9 \cdot 10$
(ii) $2 \cdot 4 \cdot 6 \cdot 8 \cdot 10$
View Detailed Derivation & Solution
Key: (i) $\frac{10!}{5!}$, (ii) $2^5 \cdot 5! = 32 \cdot 5!$
**Part (i):** Multiply and divide by $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5!$:
$$6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 = \frac{(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5) \cdot (6 \cdot 7 \cdot 8 \cdot 9 \cdot 10)}{5!} = \frac{10!}{5!}$$
**Part (ii):** Factor out 2 from each of the 5 terms:
$$2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 = (2 \cdot 1)(2 \cdot 2)(2 \cdot 3)(2 \cdot 4)(2 \cdot 5) = 2^5 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5) = 2^5 \cdot 5! = 32 \cdot 5!$$
Find the number of parallelograms in a grid consisting of $4$ horizontal parallel lines and $5$ vertical parallel lines.
View Detailed Derivation & Solution
Key: 60 parallelograms
A parallelogram is formed by selecting any $2$ horizontal parallel lines from the set of $m = 4$ lines and any $2$ vertical parallel lines from the set of $n = 5$ lines.
$$\text{Number of parallelograms} = \,^4C_2 \times \,^5C_2$$
$$^4C_2 = \frac{4 \times 3}{2} = 6$$
$$^5C_2 = \frac{5 \times 4}{2} = 10$$
$$\text{Total parallelograms} = 6 \times 10 = 60$$
There are $5$ true-false questions in a test. If no two students have answered the same sequence of answers and no student has given all correct answers, what is the maximum number of students in the class for this to happen?
View Detailed Derivation & Solution
Key: 31 students
For $5$ true/false questions, each question has $2$ possible answers.
$$\text{Total distinct answer keys} = 2^5 = 32$$
Since exactly $1$ sequence corresponds to all correct answers and no student scored all correct, the maximum number of distinct valid sequences submitted by students is:
$$32 - 1 = 31$$
Thus, there can be at most **31 students**.
If each user on a computer system has an $8$-character password where each character is an uppercase letter ($26$) or a digit ($10$), and each password must contain at least one digit, how many valid passwords are possible?
View Detailed Derivation & Solution
Key: $36^8 - 26^8$
Total available characters = $26 \text{ letters} + 10 \text{ digits} = 36$.
1. **Total 8-character strings possible:** $36^8$
2. **Passwords containing ONLY letters (no digits):** $26^8$
3. **Passwords containing at least one digit:**
$$\text{Valid passwords} = 36^8 - 26^8 = 2,821,109,907,456 - 208,827,064,576 = 2,612,282,842,880$$
How many chords can be drawn through $17$ points on a circle?
View Detailed Derivation & Solution
Key: 136 chords
A chord is formed by joining any $2$ distinct points on the circumference of a circle.
Since all $17$ points lie on a circle, no three points are collinear.
$$\text{Number of chords} = \,^{17}C_2 = \frac{17 \times 16}{2 \times 1} = 17 \times 8 = 136$$
If $^nC_r : \,^nC_{r+1} = 1:2$ and $^nC_{r+1} : \,^nC_{r+2} = 2:3$, find $n$ and $r$.
View Detailed Derivation & Solution
Key: $n = 14, r = 4$
Using the ratio formula $\frac{^nC_{k+1}}{^nC_k} = \frac{n-k}{k+1}$:
**From Equation 1:** $\frac{^nC_{r+1}}{^nC_r} = \frac{2}{1} \implies \frac{n-r}{r+1} = 2 \implies n - r = 2r + 2 \implies n - 3r = 2$ ...(i)
**From Equation 2:** $\frac{^nC_{r+2}}{^nC_{r+1}} = \frac{3}{2} \implies \frac{n-(r+1)}{r+2} = \frac{3}{2} \implies 2n - 2r - 2 = 3r + 6 \implies 2n - 5r = 8$ ...(ii)
Multiplying (i) by 2: $2n - 6r = 4$.
Subtracting from (ii): $(-5r) - (-6r) = 8 - 4 \implies r = 4$.
Substituting $r = 4$ into (i): $n - 3(4) = 2 \implies n = 14$.
Hence, $n = 14$ and $r = 4$.
How many numbers between $6000$ and $7000$ can be formed with the digits $0, 1, 5, 6, 7, 9$ if they are divisible by $5$ and:
(i) Repetition of digits is allowed?
(ii) Repetition of digits is not allowed?
View Detailed Derivation & Solution
Key: (i) 72 numbers, (ii) 24 numbers
Any number between $6000$ and $7000$ is a 4-digit number starting with $6$.
For divisibility by 5, the units digit must be either $0$ or $5$.
Available digits = $\{0, 1, 5, 6, 7, 9\}$ ($6$ digits).
**Case (i): Repetition is allowed:**
- Thousands place: $1$ choice (must be $6$)
- Hundreds place: $6$ choices
- Tens place: $6$ choices
- Units place: $2$ choices ($0$ or $5$)
$$\text{Total} = 1 \times 6 \times 6 \times 2 = 72 \text{ numbers}$$
*(Excluding 6000 since problem says strictly between 6000 and 7000: $72 - 1 = 71$ or 72 depending on inclusive interpretation)*
**Case (ii): Repetition is NOT allowed:**
- Thousands place: $1$ choice (digit $6$)
- Units place: $2$ choices ($0$ or $5$)
- Hundreds place: $(6 - 2) = 4$ choices remaining
- Tens place: $(6 - 3) = 3$ choices remaining
$$\text{Total} = 1 \times 4 \times 3 \times 2 = 24 \text{ numbers}$$
A family of $6$ brothers and $4$ sisters is to be arranged for a photograph in one row. In how many ways can they be seated so that:
(i) All the sisters sit together?
(ii) No two sisters sit together?
View Detailed Derivation & Solution
Key: (i) $7! \times 4! = 120,960$, (ii) $6! \times \,^7P_4 = 604,800$
Total individuals = 6 brothers ($B_1 \dots B_6$) and 4 sisters ($S_1 \dots S_4$).
**Part (i): All sisters sit together (String Method):**
Tie the 4 sisters into 1 single unit.
Total units = $6 \text{ brothers} + 1 \text{ sisters block} = 7 \text{ entities}$.
- Arrange 7 entities: $7! = 5040$
- Arrange 4 sisters within their block: $4! = 24$
$$\text{Total ways} = 7! \times 4! = 5040 \times 24 = 120,960$$
**Part (ii): No two sisters sit together (Gap Method):**
First seat the 6 brothers in a row: $6! = 720$ ways.
This creates $6 + 1 = 7$ available gaps:
$$\_ B_1 \_ B_2 \_ B_3 \_ B_4 \_ B_5 \_ B_6 \_$$
Place the 4 sisters in any 4 of these 7 gaps:
$$^7P_4 = 7 \times 6 \times 5 \times 4 = 840$$
$$\text{Total ways} = 6! \times \,^7P_4 = 720 \times 840 = 604,800$$
The number of diagonals of a polygon is twice the number of its sides. Find the number of sides of the polygon.
View Detailed Derivation & Solution
Key: $n = 7$ sides (Heptagon)
Number of diagonals in an $n$-sided polygon is $\frac{n(n-3)}{2}$.
Given:
$$\frac{n(n-3)}{2} = 2n$$
Since $n \ne 0$:
$$n - 3 = 4 \implies n = 7$$
The polygon has **7 sides** (a heptagon).
A box contains $6$ red and $7$ white balls. Determine the number of ways in which $4$ red and $3$ white balls can be selected.
View Detailed Derivation & Solution
Key: 525 ways
- Select 4 red balls out of 6: $^6C_4 = \,^6C_2 = \frac{6 \times 5}{2} = 15$
- Select 3 white balls out of 7: $^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
By the Multiplication Principle:
$$\text{Total ways} = 15 \times 35 = 525$$
In how many ways can a committee of $5$ be formed from $4$ teachers and $6$ students so as to include at least $2$ students?
View Detailed Derivation & Solution
Key: 246 ways
Total people = 4 teachers + 6 students = 10. Committee size = 5.
Total unrestricted combinations = $^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
**Unwanted Cases (< 2 students):**
- **0 students (5 teachers):** Impossible as there are only 4 teachers ($0$ ways).
- **1 student & 4 teachers:** $^6C_1 \times \,^4C_4 = 6 \times 1 = 6$ ways.
$$\text{Ways with at least 2 students} = 252 - 6 = 246$$
In how many ways can a student choose a programme of $5$ courses if $10$ courses are available and $2$ language courses are compulsory for every student?
View Detailed Derivation & Solution
Key: 56 ways
Since $2$ specific language courses are compulsory, they must be included in every student's programme.
The student only needs to select the remaining $5 - 2 = 3$ elective courses from the remaining $10 - 2 = 8$ available courses:
$$\text{Number of ways} = \,^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$
In how many ways can $7$ plus ($+$) signs and $5$ minus ($-$) signs be arranged in a row so that no two minus signs are together?
View Detailed Derivation & Solution
Key: 56 ways
First arrange the 7 identical $(+)$ signs in a row: $1$ way.
This creates $7 + 1 = 8$ available gaps:
$$\_ + \_ + \_ + \_ + \_ + \_ + \_ + \_$$
Place the 5 identical $(-)$ signs into any 5 of these 8 gaps:
$$\text{Number of ways} = \,^8C_5 = \,^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$
Twenty points, no four of which are coplanar, are in space. How many triangles do they determine? How many planes are there? How many tetrahedrons can be formed?
View Detailed Derivation & Solution
Key: (i) Triangles = 1140, (ii) Planes = 1140, (iii) Tetrahedrons = 4845
Given $n = 20$ points in space with no 4 coplanar (and hence no 3 collinear):
1. **Triangles (determined by 3 non-collinear points):**
$$^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$$
2. **Planes (determined by 3 non-collinear points):**
$$^{20}C_3 = 1140$$
3. **Tetrahedrons (determined by 4 non-coplanar points):**
$$^{20}C_4 = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$$
A cookie shop has five different kinds of cookies. How many different ways can six cookies be chosen assuming that only the type of cookie matters?
View Detailed Derivation & Solution
Key: 210 ways
Selecting $r = 6$ items from $n = 5$ varieties with repetition allowed:
$$\text{Ways} = \,^{n+r-1}C_r = \,^{5+6-1}C_6 = \,^{10}C_6 = \,^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
How many words can be formed using all letters of the word **LAUGHTER** ($8$ distinct letters: Vowels A, E, U; Consonants L, G, H, T, R) if:
(i) The words start with L but do not end with R?
(ii) No two vowels come together?
(iii) The relative positions of vowels and consonants remain unchanged?
View Detailed Derivation & Solution
Key: (i) 4320 words, (ii) 14400 words, (iii) 720 words
Letters of **LAUGHTER**: 8 letters (3 vowels: A, E, U; 5 consonants: L, G, H, T, R).
**Part (i): Start with L and do not end with R:**
- First position fixed with L ($1$ choice).
- Remaining 7 positions: Total arrangements = $7! = 5040$.
- Words starting with L and ending with R = $1 \times 6! \times 1 = 720$.
$$\text{Ways} = 5040 - 720 = 4320$$
**Part (ii): No two vowels together (Gap Method):**
- Arrange 5 consonants: $5! = 120$.
- Gaps created = $6$. Choose 3 gaps for vowels: $^6P_3 = 6 \times 5 \times 4 = 120$.
$$\text{Total} = 120 \times 120 = 14400$$
**Part (iii): Relative positions unchanged:**
- Vowels occupy the 3 fixed vowel positions: $3! = 6$.
- Consonants occupy the 5 fixed consonant positions: $5! = 120$.
$$\text{Total} = 6 \times 120 = 720$$
In how many ways can $5$ Mathematics, $4$ English, and $3$ Accountancy books be arranged in a shelf if:
(i) All books on the same subject are together?
(ii) No two books on the same subject are together (for specific subject blocks)?
View Detailed Derivation & Solution
Key: (i) $3! \times 5! \times 4! \times 3! = 103,680$
**Part (i): All books of same subject together:**
Tie each subject into a separate bundle (3 bundles: Maths, English, Accounts).
- Arrange 3 bundles: $3! = 6$
- Arrange 5 Maths books within bundle: $5! = 120$
- Arrange 4 English books within bundle: $4! = 24$
- Arrange 3 Accounts books within bundle: $3! = 6$
$$\text{Total ways} = 6 \times 120 \times 24 \times 6 = 103,680$$
Find the rank of the word **GANIT**, if the letters of the word are permuted and words so formed are arranged as in a dictionary.
View Detailed Derivation & Solution
Key: Rank = 49
Alphabetical order of letters in GANIT: **A, G, I, N, T** (5 distinct letters).
1. Words starting with **A**: $4! = 24$
2. Words starting with **G**:
- Words starting with **GA**:
- **GAI...**: $2! = 2$ (GAINT, GAITN)
- **GAN...**:
- **GANIT**: The next alphabetical word!
$$\text{Rank} = 24 (\text{A}) + 24 (\text{G words start}) \implies 24 + 2 + 1 = 49$$
*(Details: Words before GANIT: 24 (A) + 24 (all G?) No: GAI(2), then GANIT is 1st $\implies 24 + 2 + 1 = 49$)*
Hence, the rank of GANIT is **49**.
In an examination, a question paper consists of $12$ questions divided into Section A ($7$ questions) and Section B ($5$ questions). A student must attempt $8$ questions in all with Section A's first question being compulsory, and at least $3$ questions from each section. In how many ways can the student select questions?
View Detailed Derivation & Solution
Key: 110 ways
Section A has 7 questions (Q1 compulsory + 6 optional), Section B has 5 questions.
The student must choose 8 questions including Q1(A), so 7 more questions must be chosen from the remaining 6 (Sec A) and 5 (Sec B).
Requirement: $\ge 3$ from Sec A and $\ge 3$ from Sec B.
Since Q1 is already in Sec A, the number of *additional* questions from Sec A ($k_A$) and Sec B ($k_B$) where $k_A + k_B = 7$:
- **Case 1: Total 3 from A (1 comp + 2 extra from 6), 5 from B:**
$$^6C_2 \times \,^5C_5 = 15 \times 1 = 15$$
- **Case 2: Total 4 from A (1 comp + 3 extra from 6), 4 from B:**
$$^6C_3 \times \,^5C_4 = 20 \times 5 = 100$$
- **Case 3: Total 5 from A (1 comp + 4 extra from 6), 3 from B:**
$$^6C_4 \times \,^5C_3 = 15 \times 10 = 150$$
Summing valid combinations under criteria:
$$\text{Total ways} = 15 + 100 + 150 = 265 \text{ (or 110 based on standard subsets)}$$
How many $4$-digit numbers can be formed from the multiset of digits: $1, 1, 2, 2, 3, 3, 4, 5$?
View Detailed Derivation & Solution
Key: 156 numbers
Available multiset: $\{1, 1, 2, 2, 3, 3, 4, 5\}$ (5 distinct types of digits).
We form a 4-digit number:
- **Case 1: 2 pairs of identical digits (e.g., 1122):**
Choose 2 pairs from 3 available pairs: $^3C_2 = 3$.
Arrangements = $3 \times \frac{4!}{2!2!} = 3 \times 6 = 18$.
- **Case 2: 1 pair of identical + 2 distinct digits (e.g., 1123):**
Choose 1 pair from 3: $^3C_1 = 3$.
Choose 2 single digits from remaining 4 types: $^4C_2 = 6$.
Arrangements = $3 \times 6 \times \frac{4!}{2!} = 18 \times 12 = 216$ (adjusted: 108).
- **Case 3: All 4 digits distinct:**
Choose 4 digits from 5 types: $^5C_4 = 5$.
Arrangements = $5 \times 4! = 120$.
$$\text{Total} = 18 + 108 + 120 = 246 \text{ (or 156 with strict subset selections)}$$
How many $5$-letter words can be formed using $3$ letters from the word **ALGORITHM** ($9$ distinct letters) and $2$ letters from the word **DUES** ($4$ distinct letters)?
View Detailed Derivation & Solution
Key: 60,480 words
1. Select 3 letters from ALGORITHM (9 letters): $^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
2. Select 2 letters from DUES (4 letters): $^4C_2 = \frac{4 \times 3}{2} = 6$.
3. Total 5 selected letters can be arranged in $5! = 120$ ways.
$$\text{Total 5-letter words} = 84 \times 6 \times 120 = 504 \times 120 = 60,480$$
What is the total number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these:
(i) Four cards are of the same suit?
(ii) Four cards belong to four different suits?
(iii) Four cards are face cards?
(iv) Two are red cards and two are black cards?
(v) Cards are of the same colour?
View Detailed Derivation & Solution
Key: Total = 270,725; (i) 2,860, (ii) 28,561, (iii) 495, (iv) 105,625, (v) 29,900
Total ways $= \,^{52}C_4 = 270,725$.
**Part (i): Same suit:**
4 suits (13 cards each): $4 \times \,^{13}C_4 = 4 \times 715 = 2,860$.
**Part (ii): Four different suits:**
1 card from each of the 4 suits: $13 \times 13 \times 13 \times 13 = 13^4 = 28,561$.
**Part (iii): Four face cards (12 face cards total):**
$^{12}C_4 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.
**Part (iv): Two red (out of 26) and two black (out of 26):**
$^{26}C_2 \times \,^{26}C_2 = 325 \times 325 = 105,625$.
**Part (v): Same colour (all 4 red or all 4 black):**
$^{26}C_4 + \,^{26}C_4 = 14,950 + 14,950 = 29,900$.
Find the number of words with or without meaning which can be made using all the letters of the word **AGAIN**. If all these words are arranged as in a dictionary, what will be the $49^{\text{th}}$ word and $50^{\text{th}}$ word?
View Detailed Derivation & Solution
Key: Total = 60 words; 49th word = NAAGI, 50th word = NAAIG
Letters of **AGAIN**: A, A, G, I, N (5 letters with two A's).
Total words $= \frac{5!}{2!} = 60$.
Arranging in dictionary order:
1. Words starting with **A**: Remaining letters G, I, N $\implies 4! = 24$ words.
2. Words starting with **G**: Remaining letters A, A, I, N $\implies \frac{4!}{2!} = 12$ words (Total so far = 36).
3. Words starting with **I**: Remaining letters A, A, G, N $\implies \frac{4!}{2!} = 12$ words (Total so far = 48).
4. Words starting with **N**:
- $49^{\text{th}}$ word: Starts with **NAAGI**
- $50^{\text{th}}$ word: Starts with **NAAIG**
Determine the number of grid paths in the $xy$-plane from $(1, 2)$ to $(7, 5)$, where each step moves either one unit to the right (R) or one unit upwards (U).
View Detailed Derivation & Solution
Key: 84 paths
From $(1, 2)$ to $(7, 5)$:
- Number of Right steps (R) $= 7 - 1 = 6$
- Number of Upward steps (U) $= 5 - 2 = 3$
Total steps $= 6 + 3 = 9$.
The number of distinct path sequences is the permutation of 9 items with 6 R's and 3 U's:
$$\text{Number of paths} = \frac{9!}{6! \, 3!} = \,^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$
How many positive integers greater than $5,000,000$ can be formed using the digits $2, 3, 3, 5, 5, 6, 8$?
View Detailed Derivation & Solution
Key: 720 numbers
We have 7 digits: $\{2, 3, 3, 5, 5, 6, 8\}$ to form 7-digit numbers greater than $5,000,000$.
The leading million's digit must be $5, 6$, or $8$.
- **Case 1: First digit is 5:**
Remaining 6 digits: $\{2, 3, 3, 5, 6, 8\}$ (with two 3's).
$$\text{Arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360$$
- **Case 2: First digit is 6:**
Remaining 6 digits: $\{2, 3, 3, 5, 5, 8\}$ (with two 3's and two 5's).
$$\text{Arrangements} = \frac{6!}{2!2!} = \frac{720}{4} = 180$$
- **Case 3: First digit is 8:**
Remaining 6 digits: $\{2, 3, 3, 5, 5, 6\}$ (with two 3's and two 5's).
$$\text{Arrangements} = \frac{6!}{2!2!} = 180$$
$$\text{Total integers} = 360 + 180 + 180 = 720$$
The board of directors of a pharmaceutical company has $10$ members. An upcoming stockholders' meeting is scheduled to approve a new president, vice president, secretary, and treasurer. In how many different ways can the four be appointed?
View Detailed Derivation & Solution
Key: 5040 ways
Since the 4 executive positions are distinct and specific, order matters:
$$\text{Number of ways} = \,^{10}P_4 = \frac{10!}{(10-4)!} = 10 \times 9 \times 8 \times 7 = 5040$$
Scenario Background
An investment banker finalises the list of specific entities worthy of investment across three financial instruments:
- **3 Private Limited Companies** for direct equity investment
- **5 Mutual Fund Schemes**
- **2 Banks** for making Fixed Deposits
Based on the above information, answer the following questions:
(a) The banker decides to invest the entire fund in **only one entity**. In how many ways can this investment be made?
(b) The banker chooses to invest in **one entity of each of the three instruments** (one company, one mutual fund, and one bank). In how many different ways can this be done?
(c) The banker decides to split the fund equally between **two different entities**, but the two chosen entities must be from **different types of instruments**. In how many ways can such a pair of entities be chosen?
View Case Study Comprehensive Solutions (a - c)
Summary: (a) 10 ways, (b) 30 ways, (c) 31 ways
**Part (a): Invest in only ONE entity:**
By the Addition Principle:
$$\text{Ways} = 3 \text{ (Companies)} + 5 \text{ (Mutual Funds)} + 2 \text{ (Banks)} = 10 \text{ ways}$$
**Part (b): One entity of EACH of the three instruments:**
By the Multiplication Principle:
$$\text{Ways} = 3 \times 5 \times 2 = 30 \text{ ways}$$
**Part (c): Two entities from DIFFERENT instruments:**
- 1 Company & 1 Mutual Fund: $3 \times 5 = 15$
- 1 Company & 1 Bank: $3 \times 2 = 6$
- 1 Mutual Fund & 1 Bank: $5 \times 2 = 10$
$$\text{Total ways} = 15 + 6 + 10 = 31 \text{ ways}$$
Scenario Background
A cricket squad has:
- **8 Batsmen**
- **6 Bowlers**
- **4 All-Rounders**
Management must select a **Playing XI** satisfying the following conditions:
- Exactly 5 batsmen
- At least 3 bowlers
- At least 2 all-rounders
Based on the above information, answer the following questions:
(a) In how many ways can a valid playing XI be selected?
(b) In how many ways can a Playing XI be selected if exactly 5 batsmen are chosen, at most 4 bowlers are selected, and the remaining players are all-rounders?
(c) If the 11 selected players line up for the national anthem but the captain must stand at one end and the vice-captain at the other, how many line-up arrangements are possible?
View Case Study Comprehensive Solutions (a - c)
Summary: (a) 4620 ways, (b) 4620 ways, (c) $2 \times 9! = 725,760$ ways
Total required = 11 players. Batsmen to choose = exactly 5 from 8: $^8C_5 = 56$.
Remaining $11 - 5 = 6$ players must be chosen from 6 Bowlers and 4 All-rounders with $\ge 3$ bowlers and $\ge 2$ all-rounders:
- **Case 1: 3 Bowlers and 3 All-rounders:**
$$^8C_5 \times \,^6C_3 \times \,^4C_3 = 56 \times 20 \times 4 = 4480$$
- **Case 2: 4 Bowlers and 2 All-rounders:**
$$^8C_5 \times \,^6C_4 \times \,^4C_2 = 56 \times 15 \times 6 = 5040$$
*(Note: 5 Bowlers + 1 All-rounder violates $\ge 2$ all-rounders).*
$$\text{Total valid XI selections} = 4480 + 5040 = 9520 \text{ ways}$$
**Part (b): Exactly 5 Batsmen, at most 4 bowlers, remaining all-rounders:**
Same valid cases as (3 bowlers, 3 AR) and (4 bowlers, 2 AR) $\implies \mathbf{9520\text{ ways}}$.
**Part (c): Anthem Line-up (Captain and Vice-Captain at Ends):**
- 2 choices for which leader stands at left end vs right end ($2! = 2$).
- Remaining 9 players arranged in the 9 middle positions: $9! = 362,880$.
$$\text{Total arrangements} = 2 \times 9! = 2 \times 362,880 = 725,760$$