Chapter 6 - Permutations and Combinations

Total mice = 18. Groups = 3 (Experimental 1, Experimental 2, Control).
Size of each group = $18/3 = 6$.
Number of ways to arrange them into 3 distinct groups = $\frac{18!}{6! 6! 6!} = \frac{18!}{(6!)^3}$.

Total marbles = $6W + 5R = 11$. Draw 4.
(a) Any colour: ${}^{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{24} = 330$.
(b) 2 White and 2 Red: ${}^6C_2 \times {}^5C_2 = 15 \times 10 = 150$.
(c) All same colour: (4W) or (4R).
${}^6C_4 + {}^5C_4 = 15 + 5 = 20$.

Total selection: ${}^{16}C_{11} = {}^{16}C_5 = \frac{16 \times 15 \times 14 \times 13 \times 12}{120} = 4368$.
(i) Include 2 particular players: Select remaining $11-2=9$ from $16-2=14$.
${}^{14}C_9 = {}^{14}C_5 = \frac{14 \times 13 \times 12 \times 11 \times 10}{120} = 2002$.
(ii) Exclude 2 particular players: Select 11 from $16-2=14$.
${}^{14}C_{11} = {}^{14}C_3 = \frac{14 \times 13 \times 12}{6} = 364$.

Class XI: 20, Class XII: 20. Select 11.
Constraints: At least 5 from each.
Possible cases (XI, XII):
1. (5, 6): ${}^{20}C_5 \times {}^{20}C_6$.
2. (6, 5): ${}^{20}C_6 \times {}^{20}C_5$.
Total ways = $2 \times {}^{20}C_5 \times {}^{20}C_6$.

(i) No girls: Select 5 from 7 boys. ${}^7C_5 = 21$.
(ii) At least one boy and one girl: Total - (All Boys) - (All Girls).
Total = ${}^{11}C_5 = 462$. All Boys = 21. All Girls = 0 (need 5, have 4).
Ways = $462 - 21 = 441$.
(iii) At least 3 girls:
- 3G, 2B: ${}^4C_3 \times {}^7C_2 = 4 \times 21 = 84$.
- 4G, 1B: ${}^4C_4 \times {}^7C_1 = 1 \times 7 = 7$.
Total = $84 + 7 = 91$.

We know that ${}^nP_r = {}^nC_r \times r!$.
$840 = 35 \times r! \implies r! = \frac{840}{35} = 24$.
Since $24 = 4!$, we have $r = 4$.

Using ${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$:
${}^{15}C_8 + {}^{15}C_9 = {}^{16}C_9$.
${}^{15}C_6 + {}^{15}C_7 = {}^{16}C_7$.
Expression = ${}^{16}C_9 - {}^{16}C_7$.
Since ${}^{16}C_9 = {}^{16}C_{16-9} = {}^{16}C_7$, the value is $0$.

Each of the $r$ places can be filled in $n$ ways.
Total ways = $n \times n \times \dots \times n$ ($r$ times) = $n^r$.

Word: INTERMEDIATE (12 letters).
Vowels: I, E, E, I, A, E (6 vowels: 3 E, 2 I, 1 A).
Consonants: N, T, R, M, D, T (6 consonants: 2 T, 1 N, 1 R, 1 M, 1 D).
Arrange consonants first: $\frac{6!}{2!} = 360$.
This creates 7 gaps (_ C _ C _ C _ C _ C _ C _).
Select 6 gaps for vowels: ${}^7C_6 = 7$.
Arrange vowels in these gaps: $\frac{6!}{3! 2! 1!} = 60$.
Total ways = $360 \times 7 \times 60 = 151200$.

Total balls = 12. Draw 3. Red = 5, Others = 7.
Case 1 (2 Red, 1 Other): ${}^5C_2 \times {}^7C_1 = 10 \times 7 = 70$.
Case 2 (3 Red): ${}^5C_3 = 10$.
Total ways = $70 + 10 = 80$.

Odd digits: 1, 3, 5, 7, 9 (5 digits).
We need to form a 6-digit number.
Each of the 6 places can be filled in 5 ways.
Total numbers = $5^6 = 15625$.

Let $n$ be the number of teams.
Number of matches = ${}^nC_2 = 153$.
$\frac{n(n-1)}{2} = 153 \implies n(n-1) = 306$.
$18 \times 17 = 306$, so $n = 18$.

Arrange 6 '+' signs: 1 way (identical).
This creates 7 gaps. We need to place 4 '-' signs in these gaps.
Ways = ${}^7C_4 = \frac{7 \times 6 \times 5 \times 4}{24} = 35$.

Total ways (at least 3M, 2W):
- (3M, 3W): ${}^{10}C_3 \times {}^7C_3 = 120 \times 35 = 4200$.
- (4M, 2W): ${}^{10}C_4 \times {}^7C_2 = 210 \times 21 = 4410$.
Total = 8610.
Subtract cases where 2 particular women (W1, W2) are together:
We need to select 4 more members. W1, W2 are included (so 2W already).
- Need 3M, 1W more: ${}^{10}C_3 \times {}^5C_1 = 120 \times 5 = 600$.
- Need 4M, 0W more: ${}^{10}C_4 \times {}^5C_0 = 210 \times 1 = 210$.
Together ways = $600 + 210 = 810$.
Required ways = $8610 - 810 = 7800$.

Total balls = 9. Draw 3.
Total ways = ${}^9C_3 = 84$.
Ways with no black ball (select from 2W + 4R = 6): ${}^6C_3 = 20$.
Ways with at least one black = $84 - 20 = 64$.

The correct formula is ${}^{12}C_2 - {}^{5}C_2 + 1$. The 5 collinear points form 1 line, which must be added back.

Each of the 3 letters has 5 options. Total ways = $5 \times 5 \times 5 = 5^3 = 125$.

If $m$ particular things are together, we treat them as one unit. We select $r-m$ other things from $n-m$ in ${}^{n-m}C_{r-m}$ ways. Then arrange $(r-m+1)$ units in $(r-m+1)!$ ways, and the $m$ things internally in $m!$ ways. The given formula is incorrect.

Each of the 12 stalls can be filled by any of the 3 types of animals. Total ways = $3^{12}$.

The number of combinations of $n$ objects taken some or all at a time is $2^n - 1$ (excluding the empty set).

Red balls (4 identical): Choose 1, 2, 3, or 4 (4 ways).
Black balls (5 identical): Choose 0, 1, 2, 3, 4, or 5 (6 ways).
Total selections = $4 \times 6 = 24$.

Total 18. 9 per side. 4 fixed on Side A, 3 fixed on Side B.
Remaining guests = 11. Remaining seats: Side A (5), Side B (6).
Select 5 for Side A from 11: ${}^{11}C_5 = \frac{11!}{5! 6!}$.
Arrange 9 on Side A: $9!$. Arrange 9 on Side B: $9!$.
Total = $\frac{11!}{5! 6!} (9!)(9!)$.

Cases (Group A, Group B): (2,5), (3,4), (4,3), (5,2).
Ways = ${}^6C_2 {}^6C_5 + {}^6C_3 {}^6C_4 + {}^6C_4 {}^6C_3 + {}^6C_5 {}^6C_2$.
$= (15 \times 6) + (20 \times 15) + (15 \times 20) + (6 \times 15)$
$= 90 + 300 + 300 + 90 = 780$.

Select 3 SC from 5: ${}^5C_3$.
The remaining 9 vacancies are open to all, meaning they can be filled by any of the remaining $25-3=22$ candidates.
Correct ways = ${}^5C_3 \times {}^{22}C_9$.

(a) $3 \times 4 \times 5 = 60$.
(b) $(2^3-1)(2^4-1)(2^5-1) = 7 \times 15 \times 31 = 3255$.
(c) Total subsets - No English = $2^{12} - 2^{3+4} = 4096 - 128 = 3968$.

(a) Boys and girls alternate: There are two cases: the line starts with a boy (BGBG...) or a girl (GBGB...). Each case has $5! \times 5!$ arrangements. Total ways = $2 \times 5! \times 5!$. This matches (iv). Note that (iii) is an equivalent expression.
(b) No two girls sit together: First, arrange the 5 boys in $5!$ ways. This creates 6 gaps (_B_B_B_B_B_). To ensure no two girls are together, place the 5 girls in these 6 gaps. This can be done in ${}^6P_5 = 6!$ ways. Total ways = $5! \times 6!$. This matches (i).
(c) All the girls sit together: Treat the 5 girls as a single unit. Now we arrange 6 items (5 boys + 1 unit of girls), which can be done in $6!$ ways. The 5 girls within their unit can be arranged in $5!$ ways. Total ways = $6! \times 5!$. This also matches (i).
(d) All the girls are never together: This is the total number of arrangements minus the arrangements where all girls are together. Total arrangements = $10!$. Arrangements with girls together = $6! \times 5!$. Required ways = $10! - (6! \times 5!)$. This matches (ii).

(a) ${}^{10}C_2 \times {}^{20}C_3$. Matches (iv).
(b) ${}^9C_1 \times {}^{20}C_3$. Matches (iii).
(c) ${}^{10}C_2 \times {}^{19}C_2$. Matches (ii).
(d) ${}^{10}C_2 \times {}^{19}C_3$. Matches (i).

(a) Total numbers formed using 4 distinct digits from 7 is ${}^7P_4 = 7 \times 6 \times 5 \times 4 = 840$. Matches (i).
(b) Divisible by 2: Last digit must be even (2, 4, 6). 3 choices. Remaining 3 digits from 6: ${}^6P_3$. Total = $3 \times 120 = 360$. Matches (iii).
(c) Divisible by 25: Last two digits must be 25 or 75 (since 0 is not available for 50). 2 cases. Remaining 2 digits from 5: ${}^5P_2$. Total = $2 \times 20 = 40$. Matches (iv).
(d) Divisible by 4: Last two digits must be divisible by 4. Possible pairs from {1..7}: 12, 16, 24, 32, 36, 52, 56, 64, 72, 76 (10 pairs). For each pair, remaining 2 digits from 5: ${}^5P_2 = 20$. Total = $10 \times 20 = 200$. Matches (ii).

(a) ${}^6P_4 = 360$. Matches (iii).
(b) $6! = 720$. Matches (i).
(c) First is O or A (2). Remaining 5! $2 \times 120 = 240$. Matches (ii).