Combinations

The number of women can be (3), (4), (5), or (6). The correct sum is (\binom{8}{3}\binom{9}{3}+\binom{8}{4}\binom{9}{2}+\binom{8}{5}\binom{9}{1}+\binom{8}{6}=7756).

If exactly 3 red balls are selected, the remaining 3 of the 6 balls must be blue. The number of ways to choose 3 of the 9 red balls is \\(\binom{9}{3}=84\\), and to choose 3 of the 8 blue balls is \\(\binom{8}{3}=56\\). Therefore, the total number of selections is \\(\binom{9}{3}\binom{8}{3}=84\times56=4704\\). Option 2352 is only half of this product, so it is not correct. Exam tip: when fixed numbers are chosen from separate groups, multiply the corresponding combinations.

You can choose (3), (4), or (5) from the first (5) questions. The total is (\binom{5}{3}\binom{8}{5}+\binom{5}{4}\binom{8}{4}+\binom{5}{5}\binom{8}{3}=966).

Total ways are (\binom{15}{6}=5005) and ways with both special books are (\binom{13}{4}=715). Hence (5005-715=4290).

Choose (3) from (6) special players and (4) from the remaining (9). The ways are (\binom{6}{3}\binom{9}{4}=2520).

Total pairs are (\binom{22}{2}=231). Replacing (\binom{9}{2}) and (\binom{6}{2}) by (1), (1) for collinear groups gives (182).

Total triples are (\binom{19}{3}=969). Failed triples are (\binom{8}{3}+\binom{6}{3}=76), so (969-76=893).

This is a combinations problem because the order of the two teams in a pair does not matter. The number of distinct pairs among 18 teams is C(18, 2) = 18 × 17 ÷ 2 = 153. Each pair plays 4 matches, so the total number is 153 × 4 = 612. Therefore option D is correct. Multiplying 18 by 4 would count opponents incorrectly and ignore the pairing structure. Option A is only twice the number of pairs, option B does not apply the four-match condition correctly, and option C is another incorrect product. No pair should be counted twice.

Total ways are (\binom{13}{6}=1716) and ways with both special persons are (\binom{11}{4}=330). Hence (1716-330=1386).

With at least 4 mathematics books, the number of mathematics books selected can be 4, 5, 6, or 7. Hence the total number of selections is \(\binom{12}{4}\binom{9}{3}+\binom{12}{5}\binom{9}{2}+\binom{12}{6}\binom{9}{1}+\binom{12}{7}\binom{9}{0}\) = \(41580+28512+8316+792=79200\). Therefore, 79200 is correct. Using only \(\binom{12}{4}\binom{9}{3}\) counts exactly 4 mathematics books, but the question also allows more than 4. Exam tip: for “at least” conditions, add all the permitted cases above the minimum.

The number of English books can be (2), (3), (4), or (5). The sum of all valid cases is (108900).

The number of doctors will be (4), (5), (6), or (7). The total is (\binom{11}{4}\binom{8}{3}+\binom{11}{5}\binom{8}{2}+\binom{11}{6}\binom{8}{1}+\binom{11}{7}=35442).

The direct answer is option C, 9373. The permitted choices are exactly 4, 5, or 6 gifts. Therefore calculate \(\binom{15}{4}+\binom{15}{5}+\binom{15}{6}\). The values are \(1365+3003+5005=9373\). Option A, 8008, is smaller than the complete sum and misses valid selections. Option B, 9009, is also not the sum. Option C is the exact total. Option D, 12376, overcounts and includes choices outside the stated range or uses an incorrect calculation. Since gifts are different but selection order is irrelevant, combinations—not permutations—are required. A useful check is that the answer must be greater than the largest single term 5005 but less than the sum of many broader choices.

One color is fixed and one is excluded. Choose the remaining (5) colors from (11) and subtract (\binom{9}{5}) selections missing both other colors to get (336).

Choose (3) from (6) special students and (4) from the remaining (9). The ways are (\binom{6}{3}\binom{9}{4}=2520).

The number of special students can be (3), (4), (5), or (6). The total is (\binom{6}{3}\binom{11}{5}+\binom{6}{4}\binom{11}{4}+\binom{6}{5}\binom{11}{3}+\binom{6}{6}\binom{11}{2}=15235).

With exactly 4 men in the committee, the remaining 3 members must be women. There are \(\binom{12}{4}=495\) ways to choose 4 men and \(\binom{10}{3}=120\) ways to choose 3 women. Hence, the total number of committees is \(\binom{12}{4}\binom{10}{3}=495\times120=59400\). Using only \(\binom{22}{7}\) would be incorrect because it does not enforce the condition of exactly 4 men. Exam tip: For an “exactly” condition, choose separately from each group and multiply the results.

The number of pens can be (2), (3), (4), or (5). The total is (\binom{10}{2}\binom{8}{5}+\binom{10}{3}\binom{8}{4}+\binom{10}{4}\binom{8}{3}+\binom{10}{5}\binom{8}{2}=29736).

Using \(\binom{n}{2}=\frac{n(n-1)}{2}\), we get \(\frac{n(n-1)}{2}=190\), or \(n(n-1)=380\). Since \(20\times19=380\), \(n=20\). For comparison, \(\binom{19}{2}=171\), not 190. Exam tip: For questions involving \(\binom{n}{2}\), apply \(n(n-1)/2\) and look for consecutive factors of the resulting product.

Using the symmetry property, \(\binom{n}{r}=\binom{n}{n-r}\). For \(\binom{n}{7}=\binom{n}{11}\), we need \(11=n-7\). Thus, \(n=7+11=18\). If \(n=17\), 7 and 11 are not complementary lower indices. Exam tip: when two distinct binomial coefficients are equal, their lower indices usually add up to \(n\).

There are 2 choices for the particular object that is included. The remaining \(r-1\) objects must then be chosen from the other \(n-2\) objects, giving \(2\binom{n-2}{r-1}\). Exam tip: split “exactly one” into mutually exclusive cases.

By Pascal's identity (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the answer is (\binom{14}{7}).

Using Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\). Therefore, \(\binom{15}{7}-\binom{14}{7}=\binom{14}{6}=3003\). The value 3432 is \(\binom{14}{7}\), not the required difference. Exam tip: Use Pascal’s identity for such differences instead of expanding factorials.

The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (5) elements are chosen from (10), so (\binom{10}{5}=252).

The element (1) is fixed so choose (5) elements from (12). Subtracting (\binom{10}{3}=120) cases containing both (5), (6) gives (792-120=672).