Combinations & the Binomial Theorem
Combinations & the Binomial Theorem
An unordered selection of $r$ objects from $n$ distinct objects:
$$\binom{n}{r} = C(n,r) = \frac{n!}{r!(n-r)!}$$
$$\binom{n}{0} = \binom{n}{n} = 1 \qquad \binom{n}{r} = \binom{n}{n-r}$$
$$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r} \quad \text{(Pascal's Rule)}$$
$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$
A committee of 3 is chosen from 10 people. How many committees?
$$\binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{720}{6} = 120$$
Expand $(x+2)^4$.
$$\binom{4}{0}x^4 + \binom{4}{1}x^3(2) + \binom{4}{2}x^2(4) + \binom{4}{3}x(8) + \binom{4}{4}(16)$$
$$= x^4 + 8x^3 + 24x^2 + 32x + 16$$
How many 5-card hands from a 52-card deck?
$$\binom{52}{5} = 2{,}598{,}960$$
Practice Problems
Show Answer Key
1. $35$
2. $45$
3. $1$
4. $\binom{12}{4} = 495$
5. Both equal $15$ ✓
6. $4 + 6 = 10 = \binom{5}{2}$ ✓
7. $a^3 + 3a^2b + 3ab^2 + b^3$
8. $\binom{5}{2} \cdot 3^3 \cdot 1 = 10 \cdot 27 = 270$... wait: coeff of $x^2$ means $\binom{5}{3}3^3 = 10 \cdot 27 = 270$
9. $20$
10. $\binom{8}{3} = 56$
11. $126$
12. $28 - 8 = 20$