Combinations

Correct answer: B. 7

For an integer value of \(n\), \(\frac{(n+2)!}{(n-2)!}=(n+2)(n+1)n(n-1)\). Substituting \(n=7\) gives \(9\times8\times7\times6=3024\), so the correct answer is 7. The nearby option 8 gives \(10\times9\times8\times7=5040\), so it is incorrect. Exam tip: when dividing factorials, cancel the common initial factors and write only the four consecutive remaining factors.

Correct answer: C. 40

First, \(\frac{10!}{2!\cdot 8!}=\frac{10\times 9}{2}=45\). Therefore, \(45\times\frac{8}{9}=5\times 8=40\), so option C is correct. In an exam, cancel common factors before expanding the entire factorial to save time.

Correct answer: A. (536)

Total triples are (\binom{16}{3}=560). Failed triples are (\binom{6}{3}+\binom{4}{3}=24), so (560-24=536).

Correct answer: C. 273

First count the unordered pairs of teams using combinations, because the pair consisting of Team A and Team B is the same as the pair consisting of Team B and Team A. For 14 teams, the number of pairs is C(14,2) = 14 × 13 ÷ 2 = 91. Each pair plays 3 matches, so the total number of matches is 91 × 3 = 273. Therefore option C is correct. Option A counts only the team pairings and ignores the three matches per pair. Option B would use two matches per pairing, while option D would use four. Counting ordered pairs would also double-count each matchup, so division by 2 is essential.

Correct answer: B. (378)

Total ways are (\binom{11}{5}=462) and ways with both special persons are (\binom{9}{3}=84). Hence (462-84=378).

Correct answer: C. 10584

With at least 3 mathematics books, the number of mathematics books selected can be 3, 4, 5, or 6. Hence, the total number of selections is
\(\binom{10}{3}\binom{7}{3}+\binom{10}{4}\binom{7}{2}+\binom{10}{5}\binom{7}{1}+\binom{10}{6}\binom{7}{0}\)
\(=4200+4410+1764+210=10584\). Therefore, 10584 is correct. Counting only the case of exactly 3 mathematics books misses the other valid cases. Exam tip: for “at least” conditions, add all permissible cases.

Correct answer: A. (6006)

Total ways are (\binom{17}{5}=6188). Removing only English (\binom{8}{5}=56) and only Hindi (\binom{9}{5}=126) gives (6006).

Correct answer: C. (3612)

The number of doctors will be (4), (5), or (6). The total is (\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612).

Correct answer: C. (2288)

The direct answer is option C, 2288. “At least 3 and at most 5” means the number selected may be 3, 4, or 5. Since order does not matter, use combinations: \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}\). These are \(286+715+1287=2288\). Option A, 1716, omits one allowed group or uses an incorrect count. Option B, 2002, is not the required sum and is only one combination value here. Option C includes all three permitted sizes exactly once, so it is correct. Option D, 2717, overcounts and does not equal the required combination sum. The words “at least” give the lower bound and “at most” give the upper bound; do not include 2 or 6 gifts.

Correct answer: B. (126)

One color is fixed and one is removed so choose the remaining (4) colors from (9). The ways are (\binom{9}{4}=126).

Correct answer: C. (350)

Direct answer: Option C, 350 ways. Because a team is an unordered selection, combinations are appropriate. Exactly 3 of the 5 special students are selected, giving \(\binom{5}{3}=10\) choices. The team has 6 members, so the other \(6-3=3\) must come from the 7 non-special students. These can be chosen in \(\binom{7}{3}=35\) ways. Multiplying independent choices gives \(10\times35=350\). Option A, 280, is too small and does not equal the required product. Option B, 320, is another incorrect count. Option C, 350, is correct because it enforces exactly three special members and exactly three others. Option D, 420, is too large and overcounts. A common mistake is to choose 6 from all 12 first; that would not guarantee exactly three special students. Always split the selection into the specified groups.

Correct answer: A. (3535)

Total ways are (\binom{15}{6}=5005). Removing selections with (0) and (1) special student gives (5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535).

Correct answer: C. 6720

Since the committee must contain exactly 3 men, the remaining 3 members must be women. The number of ways to choose 3 men from 10 is \(\binom{10}{3}=120\), and the number of ways to choose 3 women from 8 is \(\binom{8}{3}=56\). Therefore, the total number of committees is \(\binom{10}{3}\binom{8}{3}=120\times56=6720\). Option 5040 results from an incorrect count for selecting the women. Exam tip: When “exactly” is specified, choose the required members from each group separately and multiply.

Correct answer: C. (10752)

The number of pens can be (2), (3), or (4). The total is (\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752).

Correct answer: C. 16

Using \(\binom{n}{2}=\frac{n(n-1)}{2}\), we get \(\frac{n(n-1)}{2}=120\), so \(n(n-1)=240\). Since \(16\times15=240\), \(n=16\). For the close distractor 15, \(\binom{15}{2}=105\), not 120. Exam tip: In \(\binom{n}{2}\) questions, look for two consecutive integers whose product is twice the given value.

Correct answer: C. 14

Use the symmetry property of combinations: \(\binom{n}{r}=\binom{n}{n-r}\). For \(\binom{n}{5}=\binom{n}{9}\), 9 must be the complement of 5. Thus, \(n-5=9\), so \(n=14\). For option 13, \(n-5=8\), so it does not satisfy the equality. Exam tip: when two combinations are equal, first check whether their lower indices are complementary.

Correct answer: A. \(\binom{n}{r}=\binom{n}{n-r}\)

Choosing \(r\) objects uniquely determines the \(n-r\) objects left out, so \(\binom{n}{r}=\binom{n}{n-r}\). In option B, the total number of available objects changes. Exam tip: use \(r\leftrightarrow n-r\) for complementary selections.

Correct answer: B. (\binom{12}{5})

By Pascal's identity (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the answer is (\binom{12}{5}).

Correct answer: C. 792

Using Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\). Hence, \(\binom{13}{6}-\binom{12}{6}=\binom{12}{5}=792\). The value 924 is \(\binom{12}{6}\) itself, not the required difference. Exam tip: simplify such differences directly with Pascal’s identity instead of evaluating both combinations separately.

Correct answer: B. (84)

The elements (1) and (2) are fixed and (3) is excluded. The remaining (3) elements are chosen from (9), so (\binom{9}{3}=84).

Correct answer: C. (336)

Total subsets are (\binom{11}{6}=462) and those containing both (4), (5) are (\binom{9}{4}=126). Hence (462-126=336).

Correct answer: A. (22800)

The direct answer is option A, 22800. From 1 to 30, the multiples of 3 are 3, 6, ..., 30, so there are \(30/3=10\). The remaining \(30-10=20\) numbers are not multiples of 3. We independently choose 3 of the first group and 2 of the second group, so the multiplication principle gives \(\binom{10}{3}\binom{20}{2}=120\times190=22800\). Option A is exactly this value. B, 11400, is half the correct result and misses valid selections. C, 19000, is not the product of the required independent choices. D, 45600, doubles the correct count and therefore overcounts. The groups are disjoint, so combinations are multiplied, not added.

Correct answer: C. 18876

Use separate combinations for the two parity groups. From 1 through 25 there are 13 odd numbers and 12 even numbers. The number of ways to choose 3 odd numbers is C(13,3) = 13 × 12 × 11 ÷ 6 = 286. The number of ways to choose 2 even numbers is C(12,2) = 12 × 11 ÷ 2 = 66. Since one selection from each group is required, multiply the independent counts: 286 × 66 = 18,876. Thus option C is correct. The alternatives are not equal to this product; they can result from omitting a parity condition, using an incorrect group size, or applying a non-combination count.

Correct answer: B. (1716)

After removing (5) special students (13) students remain. Hence there are (\binom{13}{7}=1716) ways.

Correct answer: C. 22260

The group can contain 3, 4, 5, or 6 teachers. Hence, the total number of ways is \(\binom{9}{3}\binom{11}{3}+\binom{9}{4}\binom{11}{2}+\binom{9}{5}\binom{11}{1}+\binom{9}{6}\binom{11}{0}\) = \(13860+6930+1386+84=22260\). Therefore, 22260 is correct. Option 13860 counts only the case of exactly 3 teachers, whereas the question requires at least 3 teachers. Exam tip: For ‘at least’ conditions, add all permitted cases.