Counting Principles, Permutations, and Combinations: Additional Practice (Set 2)

The condition lands on the units digit, so choose that first. To be even the number must end in $2$, $4$, $6$, or $8$, which is $4$ choices.

With the units digit used up, $8$ digits remain for the hundreds place and then $7$ for the tens place, and those counts do not depend on which even digit was chosen.

$4 \times 8 \times 7 = 224$

Halving the total $P(9,3) = 504$ to get $252$ assumes evens and odds are equally common, but the digits $1$ through $9$ hold four even digits and five odd ones. The odd count is $5 \times 8 \times 7 = 280$, and indeed $224 + 280 = 504$.

Write every integer from $0$ to $999$ as a three-digit string, padding with leading zeros, so that $5$ becomes $005$. That is $1000$ strings in all.

A string with no $7$ has $9$ choices at each of its three positions, so the strings avoiding $7$ number

$9^3 = 729.$

Subtract to get the strings containing at least one $7$.

$1000 - 729 = 271$

The two integers at the edges do not disturb this: the padded $000$ stands for $0$, which is outside the range but contains no $7$, and $1000$ itself contains no $7$. So the answer for $1$ through $1000$ is $271$.

Add up the subset counts of each odd size, using $\binom{7}{k}$ for the subsets of size $k$. The sizes do not overlap, so the counts add.

$\binom{7}{1} + \binom{7}{3} + \binom{7}{5} + \binom{7}{7} = 7 + 35 + 21 + 1 = 64$

There are $64$ such subsets, which is exactly half of the $2^7 = 128$ subsets in all. That is no accident: pairing each subset with the one obtained by adding or removing a fixed element matches every odd subset to exactly one even subset.

Exactly $3$ women means the other $2$ members are men. Choose each group separately, and neither group is ordered.

$\binom{6}{3} = 20, \qquad \binom{6}{2} = 15$

The number of ways to choose the men does not depend on which women were chosen, so multiply.

$20 \times 15 = 300$

The value $792 = \binom{12}{5}$ counts all committees of five, with no condition on the mix.

Here the complement is no easier than the direct count, since 'at least $3$ seniors' allows only two cases. Split into them and add, because a committee cannot have both exactly $3$ and exactly $4$ seniors.

With exactly $3$ seniors, choose the seniors and then the single junior.

$\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60$

With exactly $4$ seniors, all four seats go to seniors and no junior is chosen.

$\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5$

The two cases do not overlap and cover everything asked for, so the addition principle gives $60 + 5 = 65$.

A number is odd exactly when its units digit is odd, so decide the units digit first. It must be $1$, $3$, or $5$, which is $3$ choices.

The remaining four digits then fill the other four places in any order, and they are all different.

$3 \times 4! = 3 \times 24 = 72$

Check the total: the even numbers end in $2$ or $4$, giving $2 \times 4! = 48$, and $72 + 48 = 120 = 5!$, which is every arrangement of the five digits.

The leading digit carries a restriction, so choose it first. It cannot be $0$, which leaves $9$ choices from $1$ through $9$.

The remaining three positions may use $0$, but not any digit already used. After the leading digit, $9$ digits remain (the ten digits minus the one used), then $8$, then $7$.

$9 \times 9 \times 8 \times 7 = 4{,}536$

The value $P(10,4) = 5{,}040$ ignores the leading-zero rule and so counts strings such as $0123$, which are not $4$-digit numbers.

A game is decided by an ordered pair of distinct teams, the host and the visitor, since the same two teams meeting at different grounds are different games.

$P(10,2) = 10 \times 9 = 90$

The same answer comes from counting unordered pairs and doubling: there are $\binom{10}{2} = 45$ pairs of teams, and each pair meets twice, giving $2 \times 45 = 90$.

Count everything, then remove what is forbidden. The unrestricted selections number

$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220.$

A forbidden selection contains both volumes, so two of its three books are already fixed and the remaining book may be any of the other $10$ on the shelf, giving $\binom{10}{1} = 10$ forbidden selections.

$220 - 10 = 210$

The value $\binom{10}{3} = 120$ counts the selections that avoid both volumes entirely, which is stricter than the rule requires, since taking exactly one volume is allowed.

Questions $1$ and $2$ are forced, so they are not choices. Two of the eight answers are settled, leaving $6$ more to be picked from the $8$ optional questions.

The set of questions answered is unordered, so this is a combination.

$\binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$

The symmetry $\binom{8}{6} = \binom{8}{2}$ is the observation that choosing $6$ questions to answer is the same act as choosing the $2$ optional questions to skip. Without the compulsory rule the answer would be $\binom{10}{8} = 45$.

Lining books up on a shelf puts them in order, and no book is used twice, so this is a permutation of $4$ objects chosen from $10$.

$P(10,4) = \frac{10!}{(10-4)!} = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5{,}040$

The expression $\frac{10!}{4!\,6!}$ is $\binom{10}{4} = 210$, which forgets the order and so answers a different question, and $10^4$ would allow the same book to appear in several places at once.

Split into cases by which pair of subjects is used. The three cases do not overlap and cover everything, so their counts add.

A math book with a physics book gives $4 \times 3 = 12$ pairs, a math book with a chemistry book gives $4 \times 2 = 8$, and a physics book with a chemistry book gives $3 \times 2 = 6$.

$12 + 8 + 6 = 26$

Complementary counting confirms it. All pairs number $\binom{9}{2} = 36$, and the same-subject pairs number $\binom{4}{2} + \binom{3}{2} + \binom{2}{2} = 6 + 3 + 1 = 10$, so the different-subject pairs number $36 - 10 = 26$.