Combinations

Correct answer: C. 24

The governing idea is independent combination selection. One red ball must be chosen from six, giving C(6,1) = 6 possibilities. Independently, one blue ball must be chosen from four, giving C(4,1) = 4 possibilities. By the multiplication principle, the total number of pairs is C(6,1) × C(4,1) = 6 × 4 = 24. Thus option C is correct. Adding 6 and 4 gives 10, but addition does not count one choice from each group. The values 20 and 30 also do not represent the product of the two independent choices. Because the balls are distinct objects, each red-blue pair is a different selection.

Correct answer: B. 10

The selection is made only from the 5 red balls, so the number of black balls is irrelevant. Therefore, the number of ways is \(\binom{5}{2}=\frac{5\times4}{2\times1}=10\). Option 15 equals \(\binom{6}{2}\), so it does not apply to 5 red balls. Exam tip: Use \(\binom{n}{r}\) when the order of selection does not matter.

Correct answer: B. 35

Only 3 chairs are to be selected, so the order of selection does not matter. Hence, the number of ways is \(\binom{7}{3}=\frac{7!}{3!4!}=35\). The value \(210\) is \(^{7}P_{3}\), which counts arrangements where order matters, so it is not applicable here. Exam tip: Use combinations when items are selected and order is irrelevant.

Correct answer: B. (78)

The direct answer is option B, 78. A committee is a group, so the order in which the two students are selected does not matter. We therefore use a combination: \(\binom{13}{2}=\frac{13\times12}{2\times1}=78\). Option B is correct. Option A, 26, is only \(13\times2\) and has no valid selection principle here. Option C, 156, is twice 78 and would count each pair twice, once as AB and once as BA. Option D, 169, is \(13^2\), which incorrectly allows repetition and treats the two positions as ordered. Since the committee has no separate posts such as president and secretary, choosing Ravi then Sita is the same committee as choosing Sita then Ravi. Thus order must not be counted. If the question had assigned different offices, a permutation would be needed instead. Memory cue: committee or team means combination unless distinct posts are explicitly named.

Correct answer: C. 15

This is a selection problem, so the order of subjects does not matter. Hence, the number of ways is \(\binom{6}{4}=\frac{6!}{4!\,2!}=15\). The value 24 may arise from an incorrect arrangement-based calculation, but no order is involved in choosing subjects. Exam tip: use \(\binom{n}{r}\) when only selection is required.

Correct answer: C. 28

Using \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\), we get \(\binom{8}{2}=\frac{8!}{2!6!}=\frac{8\times7}{2\times1}=28\). Therefore, 28 is the correct answer. The value 56 comes from multiplying 8 and 7 without dividing by \(2!\). Exam tip: remember the shortcut \(\binom{n}{2}=\frac{n(n-1)}{2}\).

Correct answer: C. 120

The notation C(10,3) represents the number of ways to choose 3 objects from 10 when the order of the chosen objects does not matter. Use the combination formula C(n,r) = n!/[r!(n−r)!]. Substituting n = 10 and r = 3 gives C(10,3) = 10!/(3!7!) = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120. Therefore option C is correct. The value 720 is the product 10 × 9 × 8 and corresponds to P(10,3), where order is counted. Dividing by 3! is essential because the same three objects can be arranged in six orders but represent only one selection. Options A and B result from incomplete or incorrect simplification.

Correct answer: B. 1

\(\binom{9}{0}\) represents the number of ways of choosing 0 objects from 9 objects. There is exactly one way to choose nothing, so \(\binom{9}{0}=1\). The number 9 is the total number of objects, not the number of combinations. Exam tip: For every \(n\), \(\binom{n}{0}=\binom{n}{n}=1\).

Correct answer: B. 1

\(\binom{7}{7}\) means choosing all 7 objects from 7 objects. This can be done in only one way, so \(\binom{7}{7}=1\). The value 7 is the number of objects, not the number of possible selections. Exam tip: in general, \(\binom{n}{n}=1\) and \(\binom{n}{0}=1\).

Correct answer: C. 12

Using \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\), \(\binom{12}{1}=\frac{12!}{1!\,11!}=12\). Therefore, 12 is correct. The value 11 would be \(\binom{11}{1}\), not \(\binom{12}{1}\). Exam tip: The number of ways to choose one item always equals the total number of items.

Correct answer: B. 36

C(9,2) represents the number of ways to select two objects from nine when order does not matter. Use the combination formula C(n,r)=n!/[r!(n−r)!]. Thus C(9,2)=9!/(2!7!)=(9×8)/2=36. Therefore option B is correct. The result can also be understood by first counting ordered selections: there are 9 choices for the first object and 8 for the second, giving 9×8=72. Every unordered pair is counted twice in that product, once as AB and once as BA, so divide by 2 to obtain 36. Hence 72 is a permutation count, not a combination count. The values 18 and 81 arise from unrelated or incomplete calculations.

Correct answer: C. 12

\(\binom{11}{1}\) means choosing 1 object from 11 objects, so \(\binom{11}{1}=11\). Also, \(\binom{11}{0}=1\), because there is exactly one way to choose no object. Hence, \(\binom{11}{1}+\binom{11}{0}=11+1=12\). Option 11 gives only the first term and ignores \(\binom{11}{0}\). Exam tip: Remember that \(\binom{n}{0}=1\) and \(\binom{n}{1}=n\).

Correct answer: C. 14

The number of ways to choose one object from \(n\) objects is \(\binom{n}{1}=n\), since each of the \(n\) objects can be chosen separately. Hence, \(\binom{n}{1}=14\) gives \(n=14\). Option 13 is incorrect because \(\binom{13}{1}=13\). Exam tip: Remember the basic results \(\binom{n}{1}=n\) and \(\binom{n}{n}=1\).

Correct answer: A. Choosing a committee of 3 students from 8 students

In choosing a committee, the order of selected members does not matter, so the same group of 3 is counted once; hence combinations apply. In awarding ranks, changing the order changes the result, so permutations are used. Exam tip: use combinations when order is irrelevant.

Correct answer: A. They are equal

Combinations follow the symmetry property \(\binom{n}{r}=\binom{n}{n-r}\). Here, \(n=6\) and \(r=2\), so \(\binom{6}{2}=\binom{6}{6-2}=\binom{6}{4}\). In fact, both values are 15, so the first is not twice the second. Exam tip: combinations with \(r\) and \(n-r\) are always equal.

Correct answer: C. 20

\(\binom{5}{2}=\frac{5\times4}{2\times1}=10\) and \(\binom{5}{3}=\frac{5\times4\times3}{3\times2\times1}=10\). Therefore, \(\binom{5}{2}+\binom{5}{3}=10+10=20\). Option 25 is incorrect because the sum of these two combinations is 20. Exam tip: use \(\binom{n}{r}=\binom{n}{n-r}\) for a quick check; here \(\binom{5}{2}=\binom{5}{3}\).

Correct answer: B. 12

There are 3 ways to choose 1 mathematics book and 4 ways to choose 1 science book. Since these are independent selections, apply the multiplication principle: \(3 \times 4=12\). The value 24 would result from incorrectly adding or multiplying extra selections. Exam tip: for fixed selections from separate groups, multiply the relevant combinations: \(\binom{3}{1}\binom{4}{1}=12\).

Correct answer: C. 60

The number of ways to choose 2 pens is \(\binom{5}{2}=10\), and the number of ways to choose 1 pencil is \(\binom{6}{1}=6\). These choices are independent, so by the multiplication principle, the total number of selections is \(\binom{5}{2}\binom{6}{1}=10\times6=60\). Getting \(120\) would incorrectly treat the two selected pens as ordered, whereas this is a selection. Exam tip: for fixed selections from separate groups, multiply the relevant combinations.

Correct answer: A. 10

The condition requires both selected balls to be black, so only the five black balls participate in the count. Because the order of selection does not matter, this is a combinations problem rather than a permutations problem. The number of choices is C(5,2) = 5!/(2!3!) = (5 × 4)/2 = 10. Therefore option A is correct. The four white balls are irrelevant because no white ball is allowed in the selected pair. A common error is to use P(5,2) = 5 × 4 = 20, which treats the same pair as different when its order is reversed; that produces option B. Values 25 and 40 do not arise from either the appropriate combination formula or the restricted selection described in the question.

Correct answer: B. 28

This is an unordered selection, so combinations are required rather than permutations. Selecting two students from eight gives C(8,2) = 8!/[2!6!]. Expand only the necessary factors: C(8,2) = [8 × 7]/[2 × 1] = 56/2 = 28. Therefore option B is correct. The order of the two selected students does not matter: choosing Ravi and Asha is the same library team as choosing Asha and Ravi. The value 56 counts ordered arrangements of two students, while 16 and 64 do not follow from the combination formula. Thus the number of distinct teams is 28.

Correct answer: B. 6

This is a selection problem, so order does not matter. Therefore, the number of ways is \(\binom{6}{5}=\binom{6}{1}=6\). Equivalently, exactly one of the 6 students is left out, and there are 6 choices for that student. Option 5 is the number of students selected, not the number of selections. Exam tip: use \(\binom{n}{r}\) when order is not involved in a selection.

Correct answer: C. 120

Here, 3 questions are to be selected from 10 distinct questions, and the order of the selected questions does not matter. Therefore, use combinations: \(\binom{10}{3}=\frac{10!}{3!\,7!}=120\). Hence, 120 is the correct answer. A value such as \(720\) can result from counting arrangements as well, so it is not appropriate here. Exam tip: When a question asks for a selection or a set, usually use \(nCr\).

Correct answer: A. 10

Two dishes are selected from 5 dishes, and the order of selection does not matter. Therefore, the number of possible plates is \(\binom{5}{2}=\frac{5\times4}{2\times1}=10\). The value 20 would result from treating order as important, so it is not the combination count. Exam tip: when items are only selected and order is not mentioned, use \(\binom{n}{r}\).

Correct answer: B. 21

The order of teachers does not matter when forming an invigilation team, so we use combinations. The number of ways is \(\binom{7}{2}=\frac{7\times6}{2\times1}=21\). Option 42 is \(7P2\), which counts different orders separately, so it is not appropriate for selecting a team. Exam tip: use combinations for selections, committees, and teams when order is irrelevant.

Correct answer: C. 126

The governing concept is a combination because the question asks only which 4 songs are chosen; it does not assign positions or an order to them. Therefore, use 9C4 = 9!/(4!5!). Cancelling the factorials gives (9×8×7×6)/(4×3×2×1) = 3024/24 = 126. Thus option C is correct. Option A, 36, does not result from the combination formula, while option B is also an incorrect partial value. Option D, 3024, is the numerator before division and would overcount the same group many times. If the songs had to be arranged in a sequence, permutations would be required instead.