Counting Principles, Permutations, and Combinations: Practice

A meal is built by a sequence of two decisions, and the second decision offers $4$ main courses no matter which appetizer was picked. That is exactly the setting of the multiplication principle, so multiply the stage counts.

$5 \times 4 = 20$

There are $20$ meals. Adding the counts to get $9$ would answer a different question, namely how many ways there are to pick a single dish that is either an appetizer or a main.

The factorial $5!$ is the product of every whole number from $5$ down to $1$, and it counts the ways to arrange $5$ distinct objects in a row.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

The value $720$ is $6!$, one step further along.

Every book is used exactly once and the order matters, so this is an arrangement of all $6$ objects. Fill the shelf positions one at a time, watching the choices shrink as each book is placed.

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

The value $46{,}656$ is $6^6$, which would allow the same book to be used in every position.

There are $2$ slots, so the product has exactly $2$ factors, starting at $7$ and dropping by one.

$P(7,2) = 7 \times 6 = 42$

The value $21$ is $\binom{7}{2}$, which forgets the order by dividing by $2!$, and $49$ is $7^2$, which would let one object fill both slots.

Count the ordered pairs first, then forget the order. There are $P(7,2) = 7 \times 6 = 42$ ordered pairs, and each $2$-element set was counted $2! = 2$ times, once per order.

$\binom{7}{2} = \frac{42}{2} = 21$

Dividing by $2!$ is the act of forgetting the order, which is the only difference between $21$ and $42$ here.

Repeats are allowed, so every position keeps all $10$ digits available no matter what came before it. Three stages of $10$ choices multiply into a power.

$10 \times 10 \times 10 = 10^3 = 1000$

The PINs run from $000$ to $999$, which is $1000$ of them, not $999$. The value $720$ is $P(10,3)$, which would forbid a digit from repeating.

The two posts are different, so naming Ana president and Ben vice-president is a different outcome from the reverse. The order matters, and nobody repeats.

$P(6,2) = 6 \times 5 = 30$

The answer $15$ counts committees, where the two people have no titles. The same $6$ people give twice as many results here, one for each way of assigning the two distinct posts.

Do not expand both factorials. Cancel $4!$ against the tail of $6!$, which leaves only the factors above $4$.

$\frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 6 \times 5 = 30$

This is exactly $P(6,2)$, the closed form $n!/(n-k)!$ with $n = 6$ and $k = 2$.

If the three letters were all different there would be $3! = 6$ arrangements, but the two O's are identical, so swapping them changes nothing. Each distinguishable word is therefore produced $2! = 2$ times.

$\frac{3!}{2!} = \frac{6}{2} = 3$

The three arrangements are TOO, OTO, and OOT.

The prizes are different, so who gets which one matters, and every person gets exactly one prize. This is an arrangement of all $3$ prizes among the $3$ people.

$3! = 3 \times 2 \times 1 = 6$

The value $27 = 3^3$ would allow one person to collect several prizes, which the problem forbids.

The committee has no titles, so rearranging the same three people gives the same committee. Order does not matter, which is precisely what $\binom{8}{3}$ counts.

$\binom{8}{3} = \frac{P(8,3)}{3!} = \frac{336}{6} = 56$

The expression $P(8,3)$ counts ordered choices, so it is $3! = 6$ times too large here, and $8^3$ would even allow one person to fill several seats.

Use the closed form and cancel before multiplying. Two factors run down from $10$ on the top, with $2!$ underneath.

$\binom{10}{2} = \frac{10 \times 9}{2 \times 1} = \frac{90}{2} = 45$

The value $90$ is $P(10,2)$, the ordered count, which is exactly $2!$ times larger.