Permutations and Combinations Questions and Answers

Answer: B

There are 300 such numbers.

Six digits are available (0, 3, 4, 5, 9, 8), used without repetition, and a four digit number cannot start with 0.

\[ \begin{aligned} &\text{first digit: 5 choices (not 0)} \\ &\text{then } 5 \times 4 \times 3 = 60 \text{ for the rest} \\ &5 \times 60 = 300 \end{aligned} \]

Equivalently, all arrangements \(6 \times 5 \times 4 \times 3 = 360\), minus those starting with 0, \(5 \times 4 \times 3 = 60\), gives 300. 360 is not offered; 270 and the others come from miscounting the restriction on the first digit.

Answer: D

There are 136 ways.

Count the complement. There are 12 balls in all, 9 of them not white:

\[ \begin{aligned} \binom{12}{3} - \binom{9}{3} &= 220 - 84 \\ &= 136 \end{aligned} \]

Direct counting agrees: one white \(3 \times \binom{9}{2} = 108\), two white \(3 \times 9 = 27\), three white 1; total 136. 220 is the count of all selections, with no condition at all.

Answer: B

The number of ways is 10.

Each die shows at least 1, so write each face as 1 + \(y_{i}\) with \(y_{i} \ge 0\). The sum 6 becomes \(y_{1} + y_{2} + y_{3} + y_{4} = 2\). The number of non-negative solutions is

\[ \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} = 10 \]

No face exceeds 3 here, so the upper limit of 6 never matters. Checking by listing: (3,1,1,1) in 4 orders and (2,2,1,1) in 6 orders, 10 in all. 4 counts only the first pattern and 6 only the second.

Answer: E

\(n = 10\).

Use \({}^{n}P_{r} = n(n-1)\cdots(n-r+1)\). The first three factors are common to both sides and cancel:

\[ \begin{aligned} n(n-1)(n-2)(n-3)(n-4) &= 42\,n(n-1)(n-2) \\ (n-3)(n-4) &= 42 = 7 \times 6 \\ n - 3 &= 7, \quad n = 10 \end{aligned} \]

Check: \({}^{10}P_{5} = 30240\) and \({}^{10}P_{3} = 720\), and \(720 \times 42 = 30240\). The smaller values 3, 5 and 7 make \({}^{n}P_{5}\) zero or too small, and 12 gives \((9)(8) = 72\), not 42.

Answer: D

The number of arrangements is 12600.

INDEPENDENCE has 12 letters: E four times, N three times, D twice, and I, P, C once each. Fixing I first and P last leaves 10 letters, N, N, N, D, D, E, E, E, E, C, to arrange in the middle:

\[ \begin{aligned} \frac{10!}{3!\,2!\,4!} &= \frac{3628800}{6 \times 2 \times 24} \\ &= \frac{3628800}{288} = 12600 \end{aligned} \]

Each repeated letter divides out the orderings of identical copies. The other options lie close to 12600, so the division has to be done exactly; none of them equals \(\frac{10!}{3!\,2!\,4!}\).

Answer: B

The number of ways is 10.

Four tosses give \(2^{4} = 16\) equally possible sequences. The number with exactly \(k\) heads is \({}^{4}C_{k}\):

\[ {}^{4}C_{2} + {}^{4}C_{3} = 6 + 4 = 10 \]

For example, the 2-head outcomes are HHTT, HTHT, HTTH, THHT, THTH and TTHH, and the 3-head outcomes are HHHT, HHTH, HTHH and THHH.

6 counts only the two-head case and 4 only the three-head case; 12 and 8 do not correspond to any count of heads.

Answer: B

The value is \(\frac{1}{2}\).

A binomial coefficient \({}^{n}C_{r}\) is greatest at the middle value(s) of \(r\):

\[ \begin{aligned} \max {}^{5}C_{r} &= {}^{5}C_{2} = {}^{5}C_{3} = 10 \\ \max {}^{6}C_{r} &= {}^{6}C_{3} = 20 \\ \frac{10}{20} &= \frac{1}{2} \end{aligned} \]

The same \(r = 3\) gives both maxima at once, so the ratio is well defined. 2 is the ratio upside down; \(\frac{5}{6}\) and \(\frac{6}{5}\) come from comparing 5 with 6 rather than the coefficients themselves.

Answer: A

The sum is 3330.

The digits 3, 5 and 7 without repetition make \(3! = 6\) numbers: 357, 375, 537, 573, 735 and 753. Each digit stands in each place (hundreds, tens, units) exactly twice.

\[ \begin{aligned} \text{sum} &= 2(3 + 5 + 7)(100 + 10 + 1) \\ &= 2 \times 15 \times 111 = 3330 \end{aligned} \]

Adding the six numbers directly gives the same total. Most options are multiples of 111, so the factor in front decides it: six numbers, with each digit appearing twice in each place, give \(2 \times 15 = 30\) lots of 111.

Answer: D

The number of ways is 144.

Boys and girls must alternate. With 4 girls and 3 boys in a row of 7, the only alternating pattern is

G B G B G B G

A pattern starting with a boy, B G B G B G, would need as many boys as girls plus one, which fails here. The 4 girls fill their seats in \(4! = 24\) ways and the 3 boys theirs in \(3! = 6\) ways.

\[ 4! \times 3! = 24 \times 6 = 144 \]

The option 288 counts the boy-first pattern as well, which is impossible with these numbers.

Answer: B

There are 16 arrangements.

BANANA has A three times, N twice and B once. B appears only once, so the word cannot start and end with B.

• Starts and ends with A: the middle four letters are B, N, N, A, arranged in \(\frac{4!}{2!} = 12\) ways.
• Starts and ends with N: the middle four letters are B, A, A, A, arranged in \(\frac{4!}{3!} = 4\) ways.

\[ 12 + 4 = 16 \]

The option 12 counts only the A case, and 20 or 24 overcount by ignoring repeated letters in the middle.