Combinations

Either (0) or (1) special object can be chosen. The total is (\binom{3}{0}\binom{9}{4}+\binom{3}{1}\binom{9}{3}=378).

The direct answer is option A, 220. Two specified members must be included, so they occupy two of the five committee places automatically. We must select the remaining \(5-2=3\) members from the other \(14-2=12\) members. Order does not matter in a committee, so the count is \(\binom{12}{3}=\frac{12\times11\times10}{3\times2\times1}=220\). Option A is correct. Option B, 364, equals \(\binom{14}{3}\) and ignores the compulsory-inclusion condition. Option C, 1001, equals \(\binom{14}{?}\) for a different selection size and is not the required count. Option D, 2002, also overcounts and does not follow the fixed-member method. The safe method is: fix the required members, find how many places remain, find how many people remain available, and then use a combination. Memory cue: committee selection ignores order, and compulsory members reduce both numbers.

After excluding (2) special students (11) students remain. So the number of ways is (\binom{11}{4}=330).

With exactly 3 boys, the remaining 2 selected students must be girls. The number of ways to choose 3 of the 7 boys is \(\binom{7}{3}=35\), and the number of ways to choose 2 of the 5 girls is \(\binom{5}{2}=10\). Therefore, the total number of selections is \(\binom{7}{3}\binom{5}{2}=35\times10=350\). Since the order of selection does not matter, combinations—not permutations—are used. Exam tip: when a question says “exactly,” first fix the required number from each group.

The cases are (2), (3), and (4) women. The correct sum is (\binom{6}{2}\binom{8}{2}+\binom{6}{3}\binom{8}{1}+\binom{6}{4}=595).

Out of the 5 balls, exactly 2 must be red, so the remaining 3 must be blue. The number of ways to choose 2 of the 6 red balls is \(\binom{6}{2}=15\), and the number of ways to choose 3 of the 7 blue balls is \(\binom{7}{3}=35\). Hence, the total number of selections is \(\binom{6}{2}\binom{7}{3}=15\times35=525\). A value such as 315 does not correctly account for the required numbers of red and blue balls. Exam tip: when “exactly” is given, first fix the number required from each group and then multiply the independent selections.

The cases are choosing (2) or (3) from the first (3). The total is (\binom{3}{2}\binom{7}{4}+\binom{3}{3}\binom{7}{3}=147).

Total ways are (\binom{12}{4}=495) and ways with both special books are (\binom{10}{2}=45). Hence (495-45=450).

The direct answer is option B, 168. Exactly two of the three special players must be selected. Choose those special players in C2 = 3 ways. There are 11 total players and 3 special players, so 8 non-special players remain. We need five players altogether, so after choosing two special players, choose three non-special players in C3 = 56 ways. By the multiplication principle, the total is C2 × C3 = 3×56 = 168. Option A, 112, is too small and results from an incorrect combination value. Option B, 168, is correct. Option C, 224, overcounts the choices. Option D, 336, counts even more than allowed and does not enforce exactly two special players. The word “exactly” is essential: one special player is not enough, while all three are not allowed. Memory cue: exactly two special = choose two from the special group and fill every remaining place from outside it.

Total pairs are (\binom{16}{2}=120) and (6) collinear points give (1) line instead of (\binom{6}{2}). Hence (120-15+1=106).

Total triples are (\binom{13}{3}=286) and (\binom{5}{3}=10) collinear triples do not form triangles. Hence (286-10=276).

The governing concept is combinations used to count unordered pairs. A match is associated with a pair of teams, and the pair Team A–Team B is the same as Team B–Team A. Thus the number of distinct team pairs among 12 teams is C(12,2) = 12×11/2 = 66. The question states that each pair plays 2 matches, so multiply by 2: 66×2 = 132. Therefore option C is correct. Option A counts only one match per pair and omits the second match. Options B and D do not result from the correct pair count multiplied by the specified two matches. Counting ordered pairs would also double-count the identity of each pair before applying the two-match condition.

Total ways are (\binom{9}{4}=126) and ways with both special persons are (\binom{7}{2}=21). Hence (126-21=105).

The cases are (2), (3), and (4) chemistry books. The correct sum is (\binom{8}{2}\binom{5}{2}+\binom{8}{3}\binom{5}{1}+\binom{8}{4}=630).

Total ways are (\binom{14}{4}=1001). Removing only English (\binom{6}{4}=15) and only Hindi (\binom{8}{4}=70) gives (916) ways.

The cases are (3), (4), and (5) doctors. The total is (\binom{7}{3}\binom{5}{2}+\binom{7}{4}\binom{5}{1}+\binom{7}{5}=546).

The selection may contain either 2 gifts or 3 gifts. Therefore, the total number of ways is \(\binom{11}{2}+\binom{11}{3}=55+165=220\). Option 165 counts only the selections of 3 gifts and does not include selections of 2 gifts. Exam tip: Add the combinations for every permitted selection size between “at least” and “at most.”

One color is fixed and one is removed so choose the remaining (3) colors from (7). The ways are (\binom{7}{3}=35).

Direct answer: Option C, 120 ways. A team is a group, so order does not matter; therefore combinations are used. Exactly 2 of the 4 special students must be chosen, so first choose them in \(\binom{4}{2}=6\) ways. The team has 5 members, leaving 3 places. There are 10-4=6 ordinary students, and 3 of them can be chosen in \(\binom{6}{3}=20\) ways. By the multiplication rule, the total is \(\binom{4}{2}\binom{6}{3}=6\times20=120\). Option A, 90, is too small and does not equal this product. Option B, 100, also comes from an incorrect count. Option C, 120, is correct because it satisfies both conditions exactly. Option D, 160, is too large and overcounts selections. Remember: for “exactly,” choose the required number from the special group and the remaining number from everyone else, then multiply.

Total ways are (\binom{13}{6}=1716) and ways with no special student are (\binom{8}{6}=28). Hence (1716-28=1688) ways.

When a committee is chosen, only the members matter; their order does not, so it is a combination. Book arrangements and assigned posts depend on order. Exam tip: use combinations whenever selection is made without order.

Total ways are (\binom{13}{5}=1287). Removing only pens (\binom{7}{5}=21) and only pencils (\binom{6}{5}=6) gives (1260).

We use \(\binom{n}{2}=\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=78\), so \(n(n-1)=156\). Since \(13\times12=156\), \(n=13\). Although 12 is a close option, \(\binom{12}{2}=66\), not 78. Exam tip: for questions involving \(\binom{n}{2}\), first apply \(\binom{n}{2}=\frac{n(n-1)}{2}\).

Using the symmetry property of combinations, \(\binom{n}{r}=\binom{n}{n-r}\). Therefore, for \(\binom{n}{4}=\binom{n}{6}\), we need \(6=n-4\). Hence, \(n=10\). Option 9 is not correct because \(\binom{9}{4}\) and \(\binom{9}{6}=\binom{9}{3}\) are not equal. Exam tip: Two combinations with the same upper number are equal when their lower numbers add up to that upper number.

Choosing \(r\) objects from \(n\) is equivalent to deciding which \(n-r\) objects are left out, so \(\binom{n}{r}=\binom{n}{n-r}\). Option D misses the factor \((n-r)!\). Exam tip: use this complementary-selection symmetry to identify equivalent combinations.