Counting Principles, Permutations, and Combinations: Practice

Count the opposite of what you want. The group has $11$ people, so the total number of committees is

$\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330.$

The committees with no woman are made entirely of the $6$ men, and there are $\binom{6}{4} = 15$ of those. Every committee either has at least one woman or has none, so subtract.

$330 - 15 = 315$

The value $600$ comes from the tempting but wrong method of appointing one woman first ($5$ ways) and filling the other three seats freely ($\binom{10}{3} = 120$ ways). That builds a committee holding two women twice, once for each woman it could have appointed first.

The word has $7$ letters: three S's, two C's, one U, and one E, and $3 + 2 + 1 + 1 = 7$.

Tagged letters would give $7! = 5{,}040$ arrangements, and each actual word is produced $3!\,2!\,1!\,1! = 12$ times.

$\frac{7!}{3!\,2!} = \frac{5{,}040}{12} = 420$

Dividing by only the $3!$ leaves $840$, which still counts every word twice, once for each order of the two C's.

Build the hand in two stages. First choose which $2$ of the $4$ aces appear, and then choose the other $3$ cards from the $48$ non-aces, since any further ace would break the word 'exactly'.

$\binom{4}{2} = 6, \qquad \binom{48}{3} = \frac{48 \times 47 \times 46}{6} = 17{,}296$

The second stage offers the same number of choices whichever aces were picked, so the two stages multiply.

$6 \times 17{,}296 = 103{,}776$

Drawing the last three cards from all $50$ remaining cards would give $117{,}600$ and would wrongly allow a third ace into a hand that is supposed to hold exactly two.

Glue the four math books into a single block. The shelf now holds $4$ objects, the block and the three science books, and they can be ordered in $4! = 24$ ways.

Inside the block the four math books are all different, so they can be ordered in $4! = 24$ ways as well, and this number does not depend on where the block sits.

$4! \times 4! = 24 \times 24 = 576$

The value $5{,}040 = 7!$ counts every arrangement of the seven books with no condition at all.

Glue the two A's into a single block. The word now consists of $6$ objects, the AA block plus L, G, E, B, and R, and all six are different from one another.

$6! = 720$

The block needs no internal factor. The two A's are identical, so swapping them inside the block produces the very same word, which is why multiplying by $2!$ to get $1{,}440$ double counts every arrangement.

For contrast, the total number of arrangements of ALGEBRA is $7!/2! = 2{,}520$, so the arrangements whose A's are apart number $2{,}520 - 720 = 1{,}800$.

First pick the five people who form one team, which can be done in $\binom{10}{5} = 252$ ways, since picking them decides the other team automatically.

But this procedure builds every split exactly twice: once when it picks team A as the chosen five, and once when it picks team B. The overcount is the same for every split, so the division principle applies with $d = 2$.

$\frac{252}{2} = 126$

The division by $2$ is needed only because the teams are unnamed. Had one team been the home team and the other the away team, the answer would have stayed $252$.

Count the opposite. All strings number $26^4 = 456{,}976$, and the strings containing no A use only the other $25$ letters at each of the four positions.

$25^4 = 390{,}625$

Every string either contains an A or does not, so subtract.

$456{,}976 - 390{,}625 = 66{,}351$

The value $70{,}304$ comes from choosing a position for a guaranteed A ($4$ ways) and filling the other three positions freely ($26^3$ ways). That method counts a string with two A's twice, so it overshoots.

Count the arrangements you do not want, which are the ones where Ana and Ben sit together, and subtract them from the total.

All seatings number $8! = 40{,}320$. For the seatings with Ana and Ben together, glue them into a block, giving $7$ objects to arrange in $7! = 5{,}040$ ways, and $2!$ ways to order Ana and Ben inside the block.

$7! \times 2! = 5{,}040 \times 2 = 10{,}080$

Subtract the forbidden seatings from all of them.

$40{,}320 - 10{,}080 = 30{,}240$

Write the combination in closed form and set it equal to $66$.

$\binom{n}{2} = \frac{n(n-1)}{2} = 66 \quad\Longrightarrow\quad n(n-1) = 132$

Expand and solve the quadratic completely.

$n^2 - n - 132 = 0 \quad\Longrightarrow\quad (n-12)(n+11) = 0$

The roots are $n = 12$ and $n = -11$, and a count of objects cannot be negative, so $n = 12$. Check it: $\binom{12}{2} = \frac{12 \times 11}{2} = 66$, while $\binom{11}{2} = 55$ falls short.

Glue the four S's into one block. What remains to be arranged is the block itself plus one M, four I's, and two P's, which is $1 + 1 + 4 + 2 = 8$ objects.

Those $8$ objects are not all distinct: the four I's are identical and so are the two P's, so use the repeated-object formula.

$\frac{8!}{4!\,2!} = \frac{40{,}320}{24 \times 2} = \frac{40{,}320}{48} = 840$

The block needs no internal factor, because the four S's are identical and rearranging them inside the block changes nothing.

The opposite of 'at least $2$ heads' is 'fewer than $2$ heads', which means exactly $0$ heads or exactly $1$ head. Those two cases are much easier to count.

There is $1$ sequence with no head at all, and the sequences with exactly one head are decided by which flip it was, so there are $\binom{10}{1} = 10$ of them.

$1 + 10 = 11$

Subtract those from the $2^{10} = 1{,}024$ sequences in all.

$1{,}024 - 11 = 1{,}013$

Subtracting only the all-tails sequence would give $1{,}023$, which is the count for 'at least $1$ head'.

Seat the restricted person first, since the condition is about her. Maria must take one of the $4$ middle chairs, and the other five people then fill the remaining five chairs in $5! = 120$ ways.

$4 \times 5! = 4 \times 120 = 480$

Complementary counting agrees. All seatings number $6! = 720$, and the seatings with Maria at an end number $2 \times 5! = 240$, so those without her at an end number $720 - 240 = 480$.