Chapter 6 - Permutations and Combinations

We know that if ${^nC_x} = {^nC_y}$, then either $x = y$ or $x + y = n$.
Here $12 \neq 8$, so $n = 12 + 8 = 20$.

Each toss has 2 outcomes (Head or Tail).
For 6 tosses, total outcomes = $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = 64$.

We have 4 distinct digits and we need to form 4-digit numbers without repetition.
Number of ways = ${^4P_4} = 4! = 24$.

Total numbers formed = $4! = 24$.
Each digit (3, 4, 5, 6) appears in the units place equal number of times: $24/4 = 6$ times.
Sum of digits in units place = $6 \times (3 + 4 + 5 + 6) = 6 \times 18 = 108$.

Selection of 2 vowels from 4 = ${^4C_2} = 6$.
Selection of 3 consonants from 5 = ${^5C_3} = 10$.
Total selections = $6 \times 10 = 60$.
These 5 letters can be arranged in $5! = 120$ ways.
Total words = $60 \times 120 = 7200$.

Sum of digits $0+1+2+3+4+5 = 15$.
To be divisible by 3, the sum of digits must be divisible by 3.
We need to select 5 digits. The sum of all 6 is 15. If we exclude one digit, the sum of remaining 5 must be divisible by 3.
Possible exclusions:
1. Exclude 0: Sum = 15. Digits {1, 2, 3, 4, 5}. Arrangements = $5! = 120$.
2. Exclude 3: Sum = 12. Digits {0, 1, 2, 4, 5}. Arrangements = $5! - 4!$ (subtracting cases starting with 0) = $120 - 24 = 96$.
Total ways = $120 + 96 = 216$.

Let $n$ be the number of persons.
Number of handshakes = ${^nC_2} = 66$.
$\frac{n(n-1)}{2} = 66 \implies n(n-1) = 132$.
$12 \times 11 = 132$, so $n = 12$.

Total ways to choose 3 points from 12 = ${^{12}C_3} = \frac{12 \times 11 \times 10}{6} = 220$.
Points on the same line do not form a triangle. Ways to choose 3 from 7 collinear points = ${^7C_3} = \frac{7 \times 6 \times 5}{6} = 35$.
Number of triangles = $220 - 35 = 185$.

To form a parallelogram, we need 2 lines from the first set and 2 from the second.
Ways = ${^4C_2} \times {^3C_2} = 6 \times 3 = 18$.

Total players = 22.
Exclude 4 players: Remaining = 18.
Include 2 specific players: We need to select $11 - 2 = 9$ more players from the remaining $18 - 2 = 16$ players.
Number of ways = ${^{16}C_9}$.

Total 5-digit telephone numbers (0-9 allowed in all places) = $10^5 = 100,000$.
Number of telephone numbers with all distinct digits = ${^{10}P_5} = 10 \times 9 \times 8 \times 7 \times 6 = 30,240$.
Numbers with at least one digit repeated = Total - Distinct = $100,000 - 30,240 = 69,760$.

Let $m$ be men and $w$ be women. Condition: $m \ge 2$ and $w = 2m$.
Case 1: $m=2, w=4$. Ways = ${^4C_2} \times {^6C_4} = 6 \times 15 = 90$.
Case 2: $m=3, w=6$. Ways = ${^4C_3} \times {^6C_6} = 4 \times 1 = 4$.
Case 3: $m=4, w=8$. Not possible (only 6 women).
Total ways = $90 + 4 = 94$.

We have digits 0-9. We need to form a 9-digit number with distinct digits.
First place (leftmost) cannot be 0. So 9 choices (1-9).
Remaining 8 places can be filled by remaining 9 digits (including 0) in ${^9P_8}$ ways.
Total = $9 \times {^9P_8} = 9 \times 9!$.

Word: ARTICLE (7 letters).
Vowels: A, I, E (3). Consonants: R, T, C, L (4).
Positions: 1 2 3 4 5 6 7. Even positions: 2, 4, 6 (3 spots).
Arrangement of 3 vowels in 3 even places = $3! = 6$.
Arrangement of 4 consonants in remaining 4 places = $4! = 24$.
Total words = $6 \times 24 = 144$.

Green (5): At least one = $2^5 - 1 = 31$.
Blue (4): At least one = $2^4 - 1 = 15$.
Red (3): Any number (0 or more) = $2^3 = 8$.
Total combinations = $31 \times 15 \times 8 = 3720$.