Permutations and Combinations Questions and Answers

Answer: C

The number of words is 60.

VIRUS has five different letters. A three-letter word without repetition is an ordered arrangement of 3 of them, a permutation:

\[ ^{5}P_{3} = 5 \times 4 \times 3 = 60 \]

30 halves this, and 40 is not a count of arrangements of these letters. 120 is \(5!\), the number of five-letter arrangements. Counting combinations \(\binom{5}{3} = 10\) would ignore order, which words do not.

Answer: B

\(1 + \frac{\alpha}{\beta} = \) \(\frac{n+1}{r}\).

The ratio of consecutive binomial coefficients is

\[ \begin{aligned} \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} &= \frac{n - r + 1}{r} \\ 1 + \frac{\alpha}{\beta} &= \frac{r + n - r + 1}{r} = \frac{n + 1}{r} \end{aligned} \]

\(\frac{n - r + 1}{r}\) is \(\frac{\alpha}{\beta}\) alone, without the 1 added. \(\frac{n+1}{r-1}\) and \(\frac{n+1}{r+1}\) have the wrong denominator. Equivalently, the result is Pascal's rule \({}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}\) divided by \({}^{n}C_{r-1}\).

Answer: D

\(r = \) 4.

\({}^{11}P_{r}\) is the product of \(r\) consecutive integers counting down from 11:

\[ \begin{aligned} {}^{11}P_{3} &= 11 \times 10 \times 9 = 990 \\ {}^{11}P_{4} &= 990 \times 8 = 7920 \end{aligned} \]

So \(r = 4\). One more factor gives \({}^{11}P_{5} = 7920 \times 7 = 55440\), so 5, 6 and 7 are too large, and 3 gives only 990.

Answer: B

There are 96 such numbers.

Using all five digits 0, 2, 4, 6, 8 once each, there are \(5! = 120\) arrangements. Those beginning with 0 are really 4-digit numbers and must be removed; there are \(4! = 24\) of them.

\[ 5! - 4! = 120 - 24 = 96 \]

Equivalently, the first digit has 4 choices (not 0) and the rest can be arranged in \(4!\) ways: \(4 \times 24 = 96\). 120 forgets to exclude a leading zero; 64 and 88 come from other miscounts.

Answer: A

\(n(S) =\) 16.

A 5-digit number cannot begin with 0, so the first digit must be 1. Each of the other four places can be 0 or 1:

\[ n(S) = 1 \times 2^{4} = 16 \]

32 counts all \(2^{5}\) strings, including those starting with 0, which are not 5-digit numbers. 8, 24 and 64 do not come from this count.

The same reasoning gives \(9 \times 10^{4}\) for all 5-digit numbers with digits 0 to 9: the leading digit has one fewer choice than the others.

Answer: E

There are 32 subsets with an odd number of elements.

A set of \(n\) elements has \(2^{n}\) subsets, split equally between odd and even sizes, because \(\sum (-1)^{k}\,{}^{n}C_{k} = 0\). With \(n = 6\):

\[ \begin{aligned} {}^{6}C_{1} + {}^{6}C_{3} + {}^{6}C_{5} &= 6 + 20 + 6 \\ &= 32 \\ &= \frac{2^{6}}{2} \end{aligned} \]

The other options are partial counts: 12 is \({}^{6}C_{1} + {}^{6}C_{5}\), and 8, 16 and 24 do not match the sum.

Answer: D

\(\frac{8}{4}({}^{7}P_{4})\) equals \({}^{8}P_{4}\).

\[ \begin{aligned} {}^{7}P_{4} &= 7 \times 6 \times 5 \times 4 = 840 \\ \frac{8}{4} \times 840 &= 1680 = 8 \times 7 \times 6 \times 5 = {}^{8}P_{4} \end{aligned} \]

This is the identity \({}^{n}P_{r} = \frac{n}{n - r}\,{}^{n-1}P_{r}\) with \(n = 8\), \(r = 4\). For comparison, \({}^{8}P_{3} = 336\), \({}^{8}P_{5} = 6720\), \({}^{8}C_{3} = {}^{8}C_{5} = 56\).

Answer: C

The number of ways is \(\frac{25!}{(5!)^{6}}\).

First count arrangements into five labelled groups of five: choose 5 for the first, 5 of the rest for the second, and so on.

\[ \begin{aligned} &\binom{25}{5}\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5} \\ &= \frac{25!}{(5!)^{5}} \end{aligned} \]

The groups are not labelled and all have the same size, so each division has been counted 5! times, once for every ordering of the five groups. Dividing by 5! gives \(\frac{25!}{(5!)^{5}\,5!} = \frac{25!}{(5!)^{6}}\). \(\frac{25!}{(5!)^{5}}\) is the answer for distinguishable groups, which the question does not have.

Answer: A

There are 168 such integers.

Using each of 2, 4, 6, 7, 8 at most once, an integer greater than 7000 is either a four-digit number starting with 7 or 8, or any five-digit number:

\[ \begin{aligned} \text{4-digit} &: 2 \times 4 \times 3 \times 2 = 48 \\ \text{5-digit} &: 5! = 120 \\ \text{total} &= 168 \end{aligned} \]

Counting only one of the two groups, or allowing repetition, gives the other options.

Answer: C

There are 120 such numbers.

A number divisible by 10 must end in 0. With 0 fixed in the units place, the other four places are filled from the remaining five digits 1, 2, 3, 4, 5 without repetition, and the leading digit can no longer be 0:

\[ {}^{5}P_{4} = 5 \times 4 \times 3 \times 2 = 120 \]

240 and 360 count numbers ending in 0 or 5 (divisible by 5) or ignore the restriction, and the larger values count five-digit numbers in general.

Remember that 75% and 87.5% completion take two and three half-lives respectively, which is the quickest way to test such statements.