Combinations

The number of special players can be (1) or (2). The total is (\binom{3}{1}\binom{10}{5}+\binom{3}{2}\binom{10}{4}=1470).

If science students are (x) and arts students are (y), then (x+y=6) and (x=y+2), so (x=4), (y=2). The ways are (\binom{10}{4}\binom{6}{2}=3150).

After excluding (4) special students (13) students remain. So the number of ways is (\binom{13}{6}=1716).

Exactly 4 boys must be selected, so the remaining 2 students must be girls. The 4 boys can be chosen from 10 in \(\binom{10}{4}=210\) ways, and the 2 girls can be chosen from 8 in \(\binom{8}{2}=28\) ways. Therefore, the total number of selections is \(\binom{10}{4}\binom{8}{2}=210\times28=5880\). Using \(\binom{10}{6}\) alone would be incorrect because it does not enforce the condition of selecting exactly 4 boys. Exam tip: choose the required members from each group separately and multiply the results.

The number of women can be (3), (4), (5), or (6). The total is (\binom{7}{3}\binom{9}{3}+\binom{7}{4}\binom{9}{2}+\binom{7}{5}\binom{9}{1}+\binom{7}{6}=4396).

In a combination, order does not matter: choosing A, B, C is the same as choosing C, B, A. In a permutation, changing the order gives a new arrangement. Exam tip: the phrase “order does not matter” indicates combinations.

You can choose (3), (4), or (5) from the first (5) questions. The total is (\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=462).

Total ways are (\binom{14}{6}=3003) and ways with both special books are (\binom{12}{4}=495). Hence (3003-495=2508).

Direct answer: Option C, 756 ways. The order of selected players does not matter, so use combinations. Exactly 2 of the 4 special players are chosen in \(\binom{4}{2}=6\) ways. Since 6 players are needed altogether, the remaining \(6-2=4\) players must come from the 9 non-special players, because \(13-4=9\). Those choices number \(\binom{9}{4}=126\). Therefore the total is \(6\times126=756\). Option A, 504, and option B, 630, are too small and do not represent the correct product. Option C, 756, is correct because it counts exactly two special players and four others. Option D, 1008, is too large and overcounts. The phrase “exactly 2” means that selecting 3 or 4 special players is forbidden. First divide the players into special and non-special groups, then multiply the two combination counts.

Total pairs are (\binom{20}{2}=190). Replacing (\binom{8}{2}) and (\binom{5}{2}) by (1), (1) for collinear groups gives (154).

Total triples are (\binom{18}{3}=816). Failed triples are (\binom{7}{3}+\binom{5}{3}=45), so (816-45=771).

Total ways are (\binom{12}{6}=924) and ways with both special persons are (\binom{10}{4}=210). Hence (924-210=714).

The cases are (4), (5), and (6) mathematics books. The total is (\binom{11}{4}\binom{8}{2}+\binom{11}{5}\binom{8}{1}+\binom{11}{6}=13398).

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

The number of doctors will be (4), (5), or (6). The total is (\binom{10}{4}\binom{8}{2}+\binom{10}{5}\binom{8}{1}+\binom{10}{6}=8106).

The direct answer is option C, 6006. “At least 4 and at most 6” permits selecting exactly 4, 5, or 6 gifts. Because order does not matter, add the three combination counts: \(\binom{14}{4}+\binom{14}{5}+\binom{14}{6}=1001+2002+3003=6006\). Option A, 4004, omits part of the permitted selections. Option B, 5005, is also too small and does not equal the full sum. Option C includes every allowed size once and is correct. Option D, 7007, overcounts. Do not use permutations because arranging selected gifts is irrelevant; only the chosen set matters. Exam cue: translate the interval directly into a sum from the lower number through the upper number.

One color is fixed and one is excluded. Choose the remaining (5) colors from (10) and subtract (\binom{8}{5}) selections missing both other colors to get (196).

Direct answer: Option B, 560 ways. A team is an unordered group, so combinations are used. Exactly 3 of the 5 special students can be selected in \(\binom{5}{3}=10\) ways. The team has 6 members, leaving 3 places. There are \(13-5=8\) non-special students, and 3 of them can be chosen in \(\binom{8}{3}=56\) ways. Multiplication gives \(10\times56=560\). Option A, 420, is too small. Option B, 560, is correct because every counted team has exactly three special and three non-special students. Option C, 700, and option D, 840, are larger incorrect totals caused by overcounting or using the wrong group sizes. Choosing 6 from all 13 would not enforce the exact condition. Exam cue: “exactly 3” fixes both parts of the team—3 special and the remaining 3 non-special.

Total ways are (\binom{16}{7}=11440). Removing selections with (0) and (1) special student gives (11440-\binom{11}{7}-\binom{5}{1}\binom{11}{6}=8800).

With exactly 3 men in the committee, the remaining 3 members must be women. There are \(\binom{11}{3}=165\) ways to choose the men and \(\binom{9}{3}=84\) ways to choose the women. Hence, the total number of committees is \(\binom{11}{3}\binom{9}{3}=165\times84=13860\). Using only \(\binom{20}{6}\) would be incorrect because it does not enforce the condition of exactly 3 men. Exam tip: When ‘exactly’ is specified, choose from each group separately and multiply.

The number of pens can be (2), (3), (4), or (5). The total is (\binom{10}{2}\binom{9}{5}+\binom{10}{3}\binom{9}{4}+\binom{10}{4}\binom{9}{3}+\binom{10}{5}\binom{9}{2}=47502).

\(\binom{n}{2}=\frac{n(n-1)}{2}\). Hence, \(\frac{n(n-1)}{2}=153\), so \(n(n-1)=306\). Since \(18\times17=306\), \(n=18\). Option 17 is not correct because \(\binom{17}{2}=136\). Exam tip: For such questions, first use \(\binom{n}{2}=\frac{n(n-1)}{2}\) and look for consecutive integers whose product matches the result.

Use the identity \(\binom{n}{r}=\binom{n}{n-r}\). For \(\binom{n}{6}=\binom{n}{10}\), 6 and 10 must be complementary with respect to \(n\). Thus, \(n-6=10\), giving \(n=16\). For option 14, \(14-6=8\), so it does not give the stated equality. Exam tip: when \(\binom{n}{r}=\binom{n}{s}\) and \(r\ne s\), usually use \(r+s=n\).

The specified object is already fixed in the selection. Thus, choose the remaining r−1 objects from the other n−1 objects, giving \(\binom{n-1}{r-1}\). \(\binom{n-1}{r}\) instead excludes the specified object. Exam tip: fix every compulsory object first.

By Pascal's identity (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the answer is (\binom{13}{6}).