Combinations

Using Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\). Therefore, \(\binom{14}{7}-\binom{13}{7}=\binom{13}{6}=1716\). Hence, option B is correct. The value \(3432\) is \(\binom{14}{7}\), not the difference. Exam tip: For differences of adjacent binomial coefficients, apply Pascal’s identity to avoid lengthy calculation.

The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (4) elements are chosen from (9), so (\binom{9}{4}=126).

The element (1) is fixed so choose (5) elements from (11). Subtracting (\binom{9}{3}=84) cases containing both (5), (6) gives (462-84=378).

After removing (6) special students (14) students remain. Hence there are (\binom{14}{7}=3432) ways.

The number of teachers in the group can be 3, 4, 5, 6, or 7. Hence, the total number of groups is \(\binom{10}{3}\binom{12}{4}+\binom{10}{4}\binom{12}{3}+\binom{10}{5}\binom{12}{2}+\binom{10}{6}\binom{12}{1}+\binom{10}{7}\binom{12}{0}\) = \(59400+46200+16632+2520+120=124872\). Therefore, option C is correct. Using only \(\binom{10}{3}\binom{12}{4}\) counts exactly 3 teachers, not at least 3 teachers. Exam tip: for “at least” conditions, add all the valid cases.

Total ways are (\binom{19}{6}=27132). Removing all-white (\binom{9}{6}=84) and all-black (\binom{10}{6}=210) gives (26838).

Since all the prizes are identical, the order of the 5 selected candidates does not matter. Hence, the number of selections is
\(\binom{15}{5}=\frac{15!}{5!\,10!}=3003\). The value 360360 is \(^{15}P_5\), which would apply if the prizes were distinct. Exam tip: when identical items are awarded, use combinations because order is irrelevant.

Total ways are (\binom{12}{6}=924) and if no special friend goes then (\binom{8}{6}=28). Hence (924-28=896).

Choose (2) from the special fruits and (4) from the remaining (8). The ways are (\binom{5}{2}\binom{8}{4}=700).

The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (3) from the remaining (8), so (\binom{2}{1}\binom{8}{3}=112).

Chemistry is fixed and biology is excluded. Choose the remaining (4) subjects from (8) and subtract (\binom{6}{2}) selections containing both mathematics and physics to get (55).

The (3) numbers are fixed and (2) are excluded. The remaining (4) numbers are chosen from (16-3-2=11), so (\binom{11}{4}=330).

For a failed triangle (3) points are chosen from the (9) collinear points. So the number is (\binom{9}{3}).

After removing (5) broken chairs (9) good chairs remain. The ways to choose (7) are (\binom{9}{7}=36).

If both are included there are (\binom{11}{4}=330) ways and if both are excluded there are (\binom{11}{6}=462) ways. The total is (792).

You can choose (4), (5), or (6) from the last (6) questions. The total is (\binom{6}{4}\binom{8}{4}+\binom{6}{5}\binom{8}{3}+\binom{6}{6}\binom{8}{2}=1414).

The number of seniors can be (0), (1), (2), or (3). The total is (\binom{9}{0}\binom{11}{7}+\binom{9}{1}\binom{11}{6}+\binom{9}{2}\binom{11}{5}+\binom{9}{3}\binom{11}{4}=48840).

Choose (1) of the two fixed cards and (5) cards from the remaining (12). The ways are (\binom{2}{1}\binom{12}{5}=1584).

The total number of subsets of 14 distinct coins is \(2^{14}=16384\). Every odd-sized selection can be paired with an even-sized selection by including or removing one fixed coin, so the numbers of even and odd selections are equal. Hence, the number of ways to select an even number of coins is \(\frac{2^{14}}{2}=2^{13}=8192\). \(16384\) counts all selections, not only the even-sized ones. Exam tip: For \(n\geq1\) distinct objects, the number of even-sized subsets and odd-sized subsets is \(2^{n-1}\) each.

The captain is excluded and the vice-captain is fixed. The remaining (5) players are chosen from (11), so (\binom{11}{5}=462).

The number of special objects can be (0), (1), or (2). The total is (\binom{6}{0}\binom{12}{7}+\binom{6}{1}\binom{12}{6}+\binom{6}{2}\binom{12}{5}=18216).

The number of boys can be (2), (3), or (4). The total is (\binom{10}{2}\binom{9}{4}+\binom{10}{3}\binom{9}{3}+\binom{10}{4}\binom{9}{2}=23310).

The direct answer is option A, 715. Three specified members must be included in a committee of seven, so those three are already fixed. The remaining places are \(7-3=4\). The remaining people are \(16-3=13\), and we choose four of them without regard to order: \(\binom{13}{4}=715\). Hence option A is correct. Option B, 1001, is not the value of \(\binom{13}{4}\); it corresponds to a different combination. Option C, 1365, is also not the required value and overcounts the choices. Option D, 3003, is much larger and cannot result from choosing only four people from the remaining 13. The fixed three members are not selected again and their internal order is irrelevant because this is a committee. The correct chain is: seven total places, three already occupied, four places left; sixteen people, three fixed, thirteen available; choose four from thirteen. Memory cue: subtract compulsory members first, then calculate one combination.

After excluding (4) special students (14) students remain. So the number of ways is (\binom{14}{7}=3432).

Exactly 4 girls must be selected, so the remaining 3 students must be boys. The number of ways to choose 4 of the 8 girls is \(\binom{8}{4}=70\), and the number of ways to choose 3 of the 10 boys is \(\binom{10}{3}=120\). Hence, the total number of selections is \(\binom{8}{4}\binom{10}{3}=70\times120=8400\). Option 11760 does not use the required split of 4 girls and 3 boys. Exam tip: when “exactly” is specified, first fix the number chosen from each group and then multiply the combinations.