Combinations

The governing concept is combination, because a selection is made and the order in which balls are chosen does not matter. Since four balls must contain equal numbers of the two colours, exactly two red and two blue balls must be selected. The red balls can be chosen in C(6,2) = 6!/(2!4!) = 15 ways, while the blue balls can be chosen in C(5,2) = 5!/(2!3!) = 10 ways. By the multiplication principle, the total number of selections is 15 × 10 = 150. Therefore, option C is correct. Options A, B and D do not result from the required product and do not correctly count the valid two-red, two-blue selections.

The direct answer is option A, 120. Two particular members must be included, so they are already fixed in the five-member committee. We need only choose the remaining \(5-2=3\) members from the other \(12-2=10\) members. Since a committee has no order, use a combination: \(\binom{10}{3}=\frac{10\times9\times8}{3\times2\times1}=120\). Option A is correct. Option B, 210, is \(\binom{10}{4}\), so it chooses four additional members and makes a committee of six, not five. Option C, 252, is \(\binom{10}{5}\), so it chooses five additional members and is too large. Option D, 792, corresponds to choosing five from 12 and ignores the requirement that the two fixed members must be included. The key step is to remove the compulsory members and count only the free choices. Memory cue: required members are already selected; subtract them from both the total committee size and the available people.

The governing concept is restricted combinations. One particular student is excluded, so the eligible population has 10 − 1 = 9 students. A team is an unordered group: rearranging its four members does not produce a new team. Therefore the number of teams is C(9,4) = 9!/(4!5!) = (9×8×7×6)/(4×3×2×1) = 126. Thus option B is correct. The value 210 equals C(10,4), but that incorrectly allows the excluded student. The value 5040 comes from treating the team as an ordered arrangement, which counts each team many times. The value 84 does not equal the required combination, so it has no valid counting basis here.

After selecting exactly 2 girls, the remaining 3 students must be boys. The number of ways to choose 2 of the 6 girls is \(\binom{6}{2}=15\), and the number of ways to choose 3 of the 8 boys is \(\binom{8}{3}=56\). Hence, the total number of selections is \(\binom{6}{2}\binom{8}{3}=15\times56=840\). Using \(\binom{14}{5}\) would not enforce the condition of exactly 2 girls. Exam tip: When “exactly” is specified, choose the required number from each group separately and multiply.

Total selections are (\binom{12}{4}=495) and all-men selections are (\binom{7}{4}=35). Hence (495-35=460).

The cases are (2), (3), and (4) red balls. The total is (\binom{6}{2}\binom{5}{2}+\binom{6}{3}\binom{5}{1}+\binom{6}{4}=275).

Total selections are (\binom{9}{5}=126) and selections with none of the first (2) are (\binom{7}{5}=21). Thus the valid selections are (105).

Total ways are (\binom{11}{4}=330) and ways with both special books are (\binom{9}{2}=36). Hence the correct count is (294).

The two special players are already selected. The remaining (3) players are chosen from (8) in (\binom{8}{3}=56) ways.

Total pairs are (\binom{15}{2}=105) but (5) collinear points give (1) line instead of (\binom{5}{2}). Hence (105-10+1=96).

Total triples are (\binom{12}{3}=220) and (\binom{4}{3}=4) triples from collinear points do not form triangles. Hence (220-4=216).

A match-up is formed by choosing any 2 teams from 9, so the number of pairs is \(\binom{9}{2}=\frac{9\times8}{2}=36\). Since each pair plays 2 matches, the total number of matches is \(36\times2=72\). Note that 36 counts only the pairs of teams, not the total matches. Exam tip: use \(\binom{n}{2}\times\) matches per pair for such league questions.

Total ways are (\binom{8}{3}=56) and if both special persons are together the third person is chosen in (6) ways. Hence the valid count is (50).

The cases are (2), (3), and (4) mathematics books. The total is (\binom{7}{2}\binom{6}{2}+\binom{7}{3}\binom{6}{1}+\binom{7}{4}=560).

Total ways are (\binom{12}{3}=220). Removing only English (\binom{5}{3}=10) and only Hindi (\binom{7}{3}=35) gives (175) ways.

The cases are (3) doctors (1) nurse and (4) doctors. The total is (\binom{6}{3}\binom{4}{1}+\binom{6}{4}=95).

The selection can be of (2), (3), or (4) gifts. The total is (\binom{10}{2}+\binom{10}{3}+\binom{10}{4}=375).

One color is fixed and one is removed so choose the remaining (2) colors from (6). The ways are (\binom{6}{2}=15).

The direct answer is option B: 45 teams. There are 3 special students, and exactly 2 of them must be included. First choose those 2 special students: \(\binom{3}{2}=3\) ways. There are 9 total students, so 6 are not special. The team has 4 members, and after selecting 2 special students we must choose 2 ordinary students from those 6: \(\binom{6}{2}=15\) ways. By the multiplication rule, the total is \(\binom{3}{2}\binom{6}{2}=3\times15=45\). Option B is correct. Option A, 36, does not use the required two-stage selection correctly. Option C, 54, overcounts the valid choices. Option D, 63, also includes invalid teams or uses an incorrect combination. We choose exactly two special students, not one, three, or an unspecified number. Memory cue: “exactly” means choose the required number from the special group, then fill the remaining places from everyone else.

The direct answer is option A, 736. We need five students from twelve, and at least one of four special students must be included. “At least one” is easiest to count by complement: count all teams, then subtract teams containing no special student. Total teams are C5 = 12!/(5!7!) = 792. If no special student is chosen, all five students must come from the remaining 8 ordinary students, giving C5 = 56. Therefore valid teams = 792 − 56 = 736. Option A, 736, is correct. Option B, 760, results from an incorrect subtraction or combination value. Option C, 770, does not remove all teams with no special student. Option D, 792, is merely the unrestricted total and includes invalid teams. Notice that “at least one” includes teams with one, two, three, or four special students, so adding separate cases is possible but less convenient. Memory cue: for “at least one,” use total minus zero.

With exactly 3 men, the remaining 2 committee members must be women. The number of ways to choose 3 of the 7 men is \(\binom{7}{3}=35\), and the number of ways to choose 2 of the 6 women is \(\binom{6}{2}=15\). Hence, the total number of committees is \(\binom{7}{3}\binom{6}{2}=35\times15=525\). Using \(\binom{13}{5}\) would be incorrect because it does not enforce the condition of exactly 3 men. Exam tip: when fixed numbers are chosen from separate groups, multiply the respective combinations.

Total ways are (\binom{11}{4}=330). Removing only pens (\binom{6}{4}=15) and only pencils (\binom{5}{4}=5) gives (310) ways.

\(\binom{n}{2}=\frac{n(n-1)}{2}\). Hence, \(\frac{n(n-1)}{2}=66\) gives \(n(n-1)=132\). Since \(12\times 11=132\), \(n=12\). For option 11, \(\binom{11}{2}=55\), not 66. Exam tip: In such combination questions, first use \(\binom{n}{2}=\frac{n(n-1)}{2}\) and look for consecutive integers whose product matches the result.

Use the symmetry property of combinations: \(\binom{n}{r}=\binom{n}{n-r}\). Here \(\binom{n}{3}=\binom{n}{5}\) and \(3\ne5\), so \(5=n-3\). Therefore, \(n=8\). For example, for \(n=7\), \(\binom{7}{3}=35\) but \(\binom{7}{5}=21\), so it is not correct. Exam tip: when \(\binom{n}{r}\) and \(\binom{n}{s}\) are equal with \(r\ne s\), first check whether \(r+s=n\).

Choosing \(r\) objects from \(n\) is equivalent to identifying the \(n-r\) objects left out. Hence \(\binom{n}{r}=\binom{n}{n-r}\). Option B is not equal in general. Exam tip: remember this as the symmetry property of combinations.