Counting Methods - ACT Math

$210$. $\frac{7!}{4!} = \frac{5040}{24} = 210$

$\binom{5}{2} = 10$. Choose positions for ones among five spots.

1. Calculate $\frac{5!}{2!2!} = \frac{120}{4} = 30$ with repeated letters.

$10^4 = 10,000$. Each digit has 10 choices, multiply for all positions.

$\binom{52}{2} = 1,326$. Order doesn't matter when choosing cards.

$2^n$. Each element can either be included or excluded.

If one event has $m$ outcomes and another has $n$, total is $m \times n$. Sequential events multiply their outcome counts.

$\binom{5}{2} = 10$. Order doesn't matter for committee selection.

The product of all positive integers up to $n$. Calculated as $n \times (n-1) \times ... \times 2 \times 1$.

$\frac{n!}{p_1!p_2!...p_k!}$. Divide by factorials of repeated object counts.

$9 \times 9 \times 8 \times 7 = 4,536$. First digit excludes 0, others exclude previous digits.

1. Multiply clothing options using multiplication principle.

$\binom{10}{4} = 210$. Same as choosing 4 from 10 people.

1. Order matters for president and vice president roles.

$20$. $\frac{6!}{3!3!} = \frac{720}{36} = 20$

$72$. $\frac{9!}{7!} = 9 \cdot 8 = 72$

1. Using $C(5,3) = \frac{5!}{3!2!} = \frac{120}{6 \cdot 2} = 10$.

1. Stars and bars: $C(4+3-1,3-1) = C(6,2) = 15$ distributions.

$n!$. The symbol for the product of all positive integers from 1 to $n$.

1. Each element can be in or out, giving $2^4 = 16$ subsets.

1. Stars and bars: $C(3+4-1,4-1) = C(6,3) = 20$ placements.

$n!$. Each position has decreasing choices, resulting in $n!$ total arrangements.

$C(n, r) = \frac{n!}{r!(n-r)!}$. Divides by $r!$ to eliminate order since combinations ignore arrangement.

1. Calculate $4! = 4 \times 3 \times 2 \times 1 = 24$ arrangements.

1. Treat the pair as one unit: $3! \times 2! = 12$ arrangements.

1. Stars and bars: $C(2+3-1,3-1) = C(4,2) = 6$ placements.

1. Circular arrangements use $(n-1)! = 7! = 5040$ for 8 objects.

1. Apply $P(5,2) = \frac{5!}{3!} = \frac{120}{6} = 20$ permutations.

$P(n, r) = \frac{n!}{(n-r)!}$. Removes $(n-r)!$ to account for unfilled positions in permutations.