Combinations

Correct answer: B. (10)

Total distributions are (^{11}C_{2}=55). Subtract (3\times{}^{6}C_{2}=45) cases where one box has at least (5) balls, giving (10).

Correct answer: A. (792)

Direct answer: Option A, 792 subsets. A subset with exactly 5 elements is formed by choosing 5 different elements from the 12-element set. Order does not matter in a subset, so use the combination \(\binom{12}{5}\). Compute it as \(\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}=792\). Option A, 792, is therefore correct. Option B, 924, corresponds to a different combination value and is not \(\binom{12}{5}\). Option C, 990, and option D, 1188, are also incorrect numerical values. We do not multiply by arrangements because the same five elements make only one subset regardless of their order. A useful check is symmetry: \(\binom{12}{5}=\binom{12}{7}\), but neither means choosing an ordered list. Memory cue: “subset” means choose, not arrange; use \(nCr\).

Correct answer: C. 512

A set with 10 elements has \(2^{10}=1024\) subsets in total. Every even-sized subset can be paired with an odd-sized subset by toggling the membership of one fixed element, so the numbers of even- and odd-sized subsets are equal. Hence, the number of odd-sized subsets is \(1024/2=512\). \(1024\) is the total number of subsets, not just the odd-sized ones. Exam tip: for \(n\geq1\), the number of even-sized subsets and odd-sized subsets is each \(2^{n-1}\).

Correct answer: D. 37

A subset with at least 6 elements can have 6, 7, or 8 elements. Therefore, the required number is
\(\binom{8}{6}+\binom{8}{7}+\binom{8}{8}=28+8+1=37\). Option 28 counts only the subsets having exactly 6 elements, so it is not correct. Exam tip: For “at least,” add combinations from the stated size up to the maximum size.

Correct answer: B. 495

Using Vandermonde’s identity, \(\sum_{r=0}^{4}{}^{7}C_r\,{}^{5}C_{4-r}={} ^{7+5}C_4={} ^{12}C_4\). Therefore, the value is \(495\). Option 330 equals \({}^{11}C_4\), so it does not apply here. Exam tip: in this identity, add the upper indices and retain the common total lower index.

Correct answer: A. \(\binom{n}{r}=\binom{n}{n-r}\)

Choosing r objects is equivalent to deciding which n-r objects are left out; these choices correspond one-to-one. Hence \(\binom{n}{r}=\binom{n}{n-r}\). Option B is the permutation formula. Exam tip: for a complement-based choice, replace r with n-r.

Correct answer: D. 56

In the binomial expansion, the general term is \(\binom{8}{r}a^{8-r}b^r\). To obtain \(a^3b^5\), take \(r=5\); hence the coefficient is \(\binom{8}{5}=\binom{8}{3}=56\). Although one may choose the positions of either the three \(a\)'s or the five \(b\)'s, both counts are equal and refer to the same term. Exam tip: first check that the exponents add up to the binomial power; here, \(3+5=8\).

Correct answer: C. (4140)

Choose (4) winning numbers and (2) numbers from the (24) non-winning numbers. The count is (^{6}C_{4}\times{}^{24}C_{2}=4140).

Correct answer: B. (33)

The cases are (1S1C1O), (1S2C), and (2S1C). Total (2\times3\times4+2\times{}^{3}C_{2}+{}^{2}C_{2}\times3=33).

Correct answer: A. (80)

First count selections with at least (1) member from (Q): (^{9}C_{4}-^{7}C_{4}=91). Subtract (^{7}C_{2}-^{5}C_{2}=11) cases containing both members of (P), giving (80).

Correct answer: D. (3150)

Choose (2) people from each profession and multiply the results. The number is (^{5}C_{2}\times{}^{6}C_{2}\times{}^{7}C_{2}=3150).

Correct answer: B. (1782)

After fixing the required member, choose (5) from the remaining (14) and subtract cases where both special members are together, (^{12}C_{3}). The answer is (^{14}C_{5}-{}^{12}C_{3}=1782).