Combinations

Either (0) or (1) special object can be chosen. The total is (\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630).

The direct answer is option A, 715. Two members are compulsory, so they already occupy two places in the six-member committee. The remaining number of places is \(6-2=4\), and the remaining available members number \(15-2=13\). Therefore choose 4 from 13: \(\binom{13}{4}=\frac{13\times12\times11\times10}{4\times3\times2\times1}=715\). Option A is correct. Option B, 1001, is \(\binom{14}{4}\), so it uses the wrong number of available members. Option C, 1365, does not equal \(\binom{13}{4}\) and is an overcount for this problem. Option D, 3003, is also far too large and does not account correctly for the two fixed members. Since this is a committee, different orders of the same four chosen members do not create new committees. Memory cue: total members minus fixed members gives the pool; committee size minus fixed members gives the number to choose.

After excluding (3) special students (11) students remain. So the number of ways is (\binom{11}{5}=462).

Exactly 3 girls must be selected, which can be done in \(\binom{7}{3}=35\) ways. The remaining 3 students must be boys, who can be selected from 9 boys in \(\binom{9}{3}=84\) ways. Therefore, the total number of selections is \(\binom{7}{3}\binom{9}{3}=35\times84=2940\). A value such as 2520 results from using an incorrect group size or selection count. Exam tip: when the word “exactly” is used, first fix the number selected from each group and then multiply the combinations.

The cases are (2), (3), (4), and (5) women. The total is (\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752).

Each selection of r objects uniquely identifies the n-r objects left out, so the two counts are equal. For example, choosing 3 from 8 means leaving 5. Exam tip: use the complementary-selection idea.

The cases are choosing (2), (3), or (4) from the first (4). The total is (\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371).

Total ways are (\binom{13}{5}=1287) and ways with both special books are (\binom{11}{3}=165). Hence (1287-165=1122).

Direct answer: Option C, 378 ways. Since only the members selected matter, not their order, use combinations. Exactly 2 of the 3 special players are required: \(\binom{3}{2}=3\). Six players are selected altogether, so the remaining number is \(6-2=4\). There are \(12-3=9\) non-special players, and 4 can be chosen in \(\binom{9}{4}=126\) ways. Hence the multiplication rule gives \(\binom{3}{2}\binom{9}{4}=3\times126=378\). Option A, 252, misses or miscounts some valid choices. Option B, 315, is not the product required by the two independent choices. Option C, 378, correctly counts every choice with exactly two special players. Option D, 504, is too large and represents overcounting. Exam cue: subtract the special group from the total before choosing the remaining players.

Total pairs are (\binom{18}{2}=153) and (7) collinear points give (1) line instead of (\binom{7}{2}). Hence (153-21+1=133).

Total triples are (\binom{14}{3}=364) and (\binom{6}{3}=20) collinear triples do not form triangles. Hence (364-20=344).

Use combinations because each match-up is an unordered pair of distinct teams. The number of pairs among 15 teams is C(15, 2) = 15 × 14 ÷ 2 = 105. The statement says that each such pair plays 2 matches, so the total number of matches is 105 × 2 = 210. Hence option B is correct. Option A is the number of pairs and would be correct only if teams met once. Options C and D do not result from multiplying the correct pair count by the stated two matches, so they overcount or use an unrelated calculation.

Total ways are (\binom{10}{5}=252) and ways with both special persons are (\binom{8}{3}=56). Hence (252-56=196).

The cases are (3), (4), and (5) chemistry books. The total is (\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932).

Total ways are (\binom{16}{5}=4368). Removing only English (\binom{7}{5}=21) and only Hindi (\binom{9}{5}=126) gives (4221) ways.

The cases are (3), (4), and (5) doctors. The total is (\binom{8}{3}\binom{6}{2}+\binom{8}{4}\binom{6}{1}+\binom{8}{5}=1176).

The selection may contain either 3 gifts or 4 gifts. Hence, the total number of ways is \(\binom{12}{3}+\binom{12}{4}=220+495=715\). Option 495 counts only the selections of 4 gifts, so it omits the selections of 3 gifts. Exam tip: “At least 3 and at most 4” means add the cases for 3 and 4 selections.

One color is fixed and one is removed so choose the remaining (4) colors from (8). The ways are (\binom{8}{4}=70).

Direct answer: Option B, 210 ways. A team has no order, so use combinations. Choose exactly 2 from the 4 special students: \(\binom{4}{2}=6\). The team needs 5 students, so choose the remaining 3 from the non-special students. Their number is \(11-4=7\), and the choices are \(\binom{7}{3}=35\). Therefore the total is \(6\times35=210\). Option A, 175, is not the product of the required choices and undercounts. Option B, 210, is correct because every selection contains exactly two special and three non-special students. Option C, 245, gives an incorrect total. Option D, 280, is too large and overcounts. The key idea is that “exactly 2” excludes selections with 0, 1, 3, or 4 special students. Memory cue: required special choices first, leftover team places from the ordinary group second, then multiply.

Total ways are (\binom{14}{6}=3003) and ways with no special student are (\binom{10}{6}=210). Hence (3003-210=2793) ways.

Exactly 2 men must be selected, so the remaining 4 committee members must be women. The number of ways to choose 2 of the 9 men is \(\binom{9}{2}=36\), and the number of ways to choose 4 of the 7 women is \(\binom{7}{4}=35\). Hence, the total number of committees is \(\binom{9}{2}\binom{7}{4}=36\times35=1260\). Therefore, option C is correct. Exam tip: When “exactly” is specified, choose the required members from each group separately and multiply the results.

Total ways are (\binom{15}{5}=3003). Removing only pens (\binom{8}{5}=56) and only pencils (\binom{7}{5}=21) gives (2926).

There are \(\binom{8}{3}\) total teams. In every invalid team, both particular students are already chosen, so the third member is selected from the remaining 6 students: \(\binom{6}{1}\). Hence subtract \(\binom{6}{1}\). Exam tip: count forbidden cases separately before subtracting.

Use the symmetry property of combinations: \(\binom{n}{r}=\binom{n}{n-r}\). For \(\binom{n}{5}=\binom{n}{7}\), we need \(7=n-5\). Thus, \(n=5+7=12\). Hence, 12 is the correct option. For example, when \(n=11\), \(\binom{11}{5}\) and \(\binom{11}{7}=\binom{11}{4}\) are not equal. Exam tip: for equal binomial coefficients, check whether the lower indices are complementary, i.e. \(r\) and \(n-r\).

A set with \(n\) elements has \(2^n\) total subsets. Exactly one of them is the empty set, so subsets with at least one element are \(2^n-1\). Exam tip: subtract 1 whenever the empty subset is excluded.