Counting Principles, Permutations, and Combinations: Additional Practice (Set 1)

The word has $6$ letters: three P's, two E's, and one R.

Tagged letters would give $6! = 720$ arrangements, and each actual word is produced $3!\,2!\,1! = 12$ times.

$\frac{6!}{3!\,2!\,1!} = \frac{720}{12} = 60$

Divide by the factorials of the repeat counts. Dividing by $3 \times 2 = 6$ instead would leave $120$, which is not a count of anything.

The toppings all go on the same pizza, so the order in which they are named changes nothing. This is an unordered choice of $3$ from $10$.

$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$

The value $720$ is $P(10,3)$, which would treat mushroom-onion-pepper as different from pepper-onion-mushroom.

Build it in two stages. Choose the committee, which is unordered, and then choose the chair from within it.

$\binom{9}{4} = 126, \qquad \text{then } 4 \text{ ways to name the chair}$

The second stage offers $4$ choices whichever committee was formed, so the stages multiply.

$126 \times 4 = 504$

Building it the other way round confirms the answer: pick the chair first ($9$ ways), then the other three members from the remaining $8$ people ($\binom{8}{3} = 56$ ways), giving $9 \times 56 = 504$.

Handle the restricted position first. The opening letter must be one of the $5$ vowels.

After that letter is used, no repeats are allowed, so the remaining three positions draw from the $25$ unused letters with shrinking counts.

$5 \times 25 \times 24 \times 23 = 5 \times 13{,}800 = 69{,}000$

The value $87{,}880 = 5 \times 26^3$ would allow the vowel to appear again later, which the no-repeat rule forbids.

The element $1$ is forced in, so it is not a decision at all. The only freedom left is what to do with the other $7$ elements, each of which is independently either in the subset or out of it.

$2^7 = 128$

A cleaner way to see it: exactly half of all $2^8 = 256$ subsets contain the element $1$, because removing or adding $1$ pairs off the subsets that contain it with the subsets that do not.

Every choice of three of the points really does give a triangle, because a line meets a circle at most twice, so no three of the points can be collinear.

A triangle is decided entirely by the set of three vertices it uses, and naming those vertices in a different order gives the same triangle. So the order does not matter.

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

The value $336 = P(8,3)$ counts ordered triples of vertices, and each triangle appears in $3! = 6$ of them, which is exactly the factor that the division removes.

The word has $8$ letters, and the only repeat is the letter E, which appears twice. Every other letter (G, O, M, T, R, Y) appears once.

Tagged letters would give $8! = 40{,}320$ arrangements, and each actual word is produced $2!$ times, once for each order of the two E's.

$\frac{8!}{2!} = \frac{40{,}320}{2} = 20{,}160$

There are $9$ people in all, so the total number of committees is $\binom{9}{4} = 126$. Count the committees you do not want, which have fewer than $2$ girls.

With no girls, all four seats go to boys: $\binom{5}{4} = 5$. With exactly one girl, choose her and then three boys: $\binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40$.

$5 + 40 = 45$

Subtract the unwanted committees from the total.

$126 - 45 = 81$

A direct count agrees: $\binom{4}{2}\binom{5}{2} + \binom{4}{3}\binom{5}{1} + \binom{4}{4}\binom{5}{0} = 60 + 20 + 1 = 81$.

Count all the arrangements, then remove the ones where the C's touch. The word has $6$ letters with the C repeated twice, so all arrangements number

$\frac{6!}{2!} = \frac{720}{2} = 360.$

For the arrangements with the C's together, glue them into a block. That leaves $5$ objects (the CC block plus I, R, L, E), all different, so there are $5! = 120$ of them. No internal factor is needed, since the two C's are identical.

$360 - 120 = 240$

There are $5$ flags with two types, three of one colour and two of the other. Use the repeated-object formula.

$\frac{5!}{3!\,2!} = \frac{120}{6 \times 2} = 10$

Seeing it as a choice makes it even clearer: the arrangement is decided entirely by which $2$ of the $5$ positions hold the blue flags, and $\binom{5}{2} = 10$. With only two types, the arrangement formula and the binomial coefficient are the same number.

A digit $0$ can never appear, because in an increasing number it would have to come first, and a $3$-digit number cannot start with $0$. So the three digits are chosen from $1$ through $9$.

Once a set of three different digits is chosen, exactly one of its $3! = 6$ orderings is increasing, so each set gives exactly one number. Choosing the set is all the work there is.

$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$

The value $504 = P(9,3)$ counts all the orderings, but only one ordering per set is increasing, which is exactly the division being performed.

A route is completely described by the order of its $7$ moves, four of them R and three of them U, so a route is an arrangement of the letters RRRRUUU.

The moves of each type are identical, so use the repeated-object formula.

$\frac{7!}{4!\,3!} = \frac{5{,}040}{24 \times 6} = \frac{5{,}040}{144} = 35$

Equivalently, a route is decided by choosing which $3$ of the $7$ moves are the up moves, and $\binom{7}{3} = 35$. The two views agree, as they must, since arranging two kinds of object is the same as choosing the positions of one of them.