Permutations and Combinations Questions and Answers

Answer: E

\({}^{n}P_{3}\) is 336.

The paper's \(n_{P_{5}}\) means \({}^{n}P_{5}\). Since \({}^{n}P_{r} = n \cdot {}^{n-1}P_{r-1}\):

\[ \begin{aligned} n &= \frac{{}^{n}P_{5}}{{}^{n-1}P_{4}} = \frac{6720}{840} = 8 \\ {}^{8}P_{3} &= 8 \times 7 \times 6 = 336 \end{aligned} \]

Check: \({}^{8}P_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 6720\). The other options are not products of three consecutive integers starting from 8.

Answer: C

The relation is \(b^{2} = a(b + c)\).

Use \({}^{n}P_{r} = {}^{n}P_{r-1} \times (n - r + 1)\) and \({}^{n}P_{r+1} = {}^{n}P_{r} \times (n - r)\). From the equal ratios,

\[ \begin{aligned} \frac{b}{a} &= \frac{{}^{n}P_{r}}{{}^{n}P_{r-1}} = n - r + 1 \\ \frac{c}{b} &= \frac{{}^{n}P_{r+1}}{{}^{n}P_{r}} = n - r \end{aligned} \]

Subtracting, \(\frac{b}{a} - \frac{c}{b} = 1\). Multiply through by \(ab\): \(b^{2} - ac = ab\), so \(b^{2} = ab + ac = a(b + c)\).

A numerical check: with \(n = 5\), \(r = 2\) the permutations are 5, 20 and 60, and \(20^{2} = 400 = 5(20 + 60)\), while for example \(c^{2} = 3600\) is not \(b(a + c) = 1300\). The other options fail this check.

Answer: C

There are 46 ways.

With at least two men and exactly twice as many women, the committee must be either 2 men and 4 women or 3 men and 6 women.

\[ \begin{aligned} \binom{3}{2}\binom{6}{4} &= 3 \times 15 = 45 \\ \binom{3}{3}\binom{6}{6} &= 1 \times 1 = 1 \\ \text{total} &= 45 + 1 = 46 \end{aligned} \]

45 is the first case alone, which forgets the committee of all nine people. A single man with two women is excluded by the condition 'at least two men'.

Answer: D

There were 8 men.

Let there be \(m\) men. Games among the men number \(\binom{m}{2}\), and games between men and the two women number \(2m\).

\[ \begin{aligned} \frac{m(m - 1)}{2} - 2m &= 12 \\ m^{2} - m - 4m &= 24 \\ m^{2} - 5m - 24 &= 0 \\ (m - 8)(m + 3) &= 0 \end{aligned} \]

So \(m = 8\). Check: 28 games among the men, 16 against women, and \(28 - 16 = 12\). With 6 men the counts would be 15 and 12, a difference of only 3.

Answer: C

The number is 105.

The digits are taken without repetition (with repetition the count is 168, which is not offered). An even number ends in 0, 2, 4 or 6, and the hundreds digit cannot be 0, so split by the last digit.

\[ \begin{aligned} \text{ends in 0} &: 6 \times 5 = 30 \\ \text{ends in 2, 4, 6} &: 3 \times (5 \times 5) = 75 \\ \text{total} &= 30 + 75 = 105 \end{aligned} \]

When the last digit is 2, 4 or 6, the hundreds digit has 5 choices (not 0, not the last digit) and the tens digit 5 (any remaining digit, including 0). 30 and 75 are the two cases separately.

Answer: A

The number of ways is 1287.

If three particular players must be included, 8 more places are filled from the remaining 13 players.

\[ \begin{aligned} \tbinom{13}{8} &= \tbinom{13}{5} \\ &= \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{120} \\ &= 1287 \end{aligned} \]

Choosing 11 from 16 without the condition would give 4368. The other values do not equal any natural count for this selection.

Answer: D

The number of ways is 441.

With at least three women among six members, count each possible split of women and men (7 women, 4 men).

\[ \begin{aligned} 3W, 3M &: \tbinom{7}{3}\tbinom{4}{3} = 35 \times 4 = 140 \\ 4W, 2M &: \tbinom{7}{4}\tbinom{4}{2} = 35 \times 6 = 210 \\ 5W, 1M &: \tbinom{7}{5}\tbinom{4}{1} = 21 \times 4 = 84 \\ 6W &: \tbinom{7}{6} = 7 \end{aligned} \]

The total is \(140 + 210 + 84 + 7 = 441\). Leaving out the all-women committee gives 434, and the larger options count committees with fewer than three women.

Answer: C

There are 45 such subsets.

A two-element subset is a choice of 2 elements from 10, with order not mattering:

\[ \binom{10}{2} = \frac{10 \times 9}{2} = 45 \]

90 counts ordered pairs, so each subset is counted twice. 20 and 40 do not come from any correct count.