Combinations

The 3 special questions cannot be selected, so they are excluded from the available questions. Thus, 5 questions must be chosen from \(15-3=12\) questions: \(\binom{12}{5}=792\). Therefore, option B is correct. The value 1365 equals \(\binom{15}{4}\), which does not follow the given selection condition. Exam tip: For combinations with forbidden items, first subtract the forbidden items from the total, then apply the combination formula.

This is a restricted-combination problem. Let the required poster be P. Since P must be included, one of the two positions in the selected pair is already fixed; only one further poster needs to be chosen. Of the seven posters, six are different from P, so the number of possible choices for the second poster is C(6,1) = 6. Hence option B is correct. Another equivalent method is to count all pairs containing P: each such pair is determined uniquely by its other poster. The answer is not 7 because P is not chosen again as an additional poster. It is not 12 because order is irrelevant: the pair {P,Q} is the same selection as {Q,P}. This distinction separates combinations from permutations.

Exclude the particular prize that must not be displayed. This leaves 8 prizes, and we need to select 3 of them. Since the order of selected prizes does not matter, the number of ways is \(\binom{8}{3}=56\). The value 84 would result from incorrectly choosing 3 prizes from all 9 prizes. Exam tip: use combinations, \(\binom{n}{r}\), when order is not important.

The phrase “all must be black” restricts the eligible objects before counting begins. The five white balls cannot appear in any valid selection, so only the seven black balls matter. We must choose four of those seven balls, and their order does not matter because the result is a selection rather than an arrangement. Therefore the required number is C(7,4) = 7!/(4!3!) = (7×6×5)/(3×2×1) = 35. Thus option B is correct. Choosing C(12,4) would ignore the colour restriction and count invalid groups containing white balls. Using a permutation would also be unsuitable, since rearranging the same four selected balls does not create a different selection.

The condition excludes every mathematics book, so all 3 books must be chosen from the 6 English books. Hence, the number of selections is \(\binom{6}{3}=\frac{6\times5\times4}{3\times2\times1}=20\). Option 15 is \(\binom{6}{2}\), which counts selections of only 2 books. Exam tip: When a category is completely excluded, remove it before forming the combination.

First exclude the particular key that must not be selected. This leaves 7 keys, and the number of ways to choose 2 of them is \(\binom{7}{2}=\frac{7\times6}{2}=21\). Using \(\binom{8}{2}=28\) would incorrectly include selections containing the prohibited key. Exam tip: for a “must not be selected” condition, remove that object before forming the combination.

Total ways are (\binom{10}{3}=120), and ways with both particular photos are (\binom{8}{1}=8). Thus (120-8=112) ways.

Because one particular candidate must be included, that candidate is fixed in every valid group. We therefore need to choose only the remaining 2 members from the other 7 candidates. Since selection order does not matter, use a combination: C(7,2) = 7 × 6 / 2 = 21. Thus there are 21 possible selections, so option B is correct. Counting C(8,3) would include groups that omit the required candidate and would violate the condition. Permutations are also inappropriate because the selected group has no order. The fixed-inclusion condition reduces the problem to choosing two people from seven, which is exactly what the calculation represents.

Answer: option C, 9. Combinations have the symmetry property nCr = nC(n-r). This means that choosing r objects from n objects gives the same number of groups as choosing the n-r objects left out. Here n = 14 and r = 5, so ¹⁴C₅ = ¹⁴C(14-5) = ¹⁴C₉. The equation also has the obvious possibility s = 5, but the question specifically says s ≠ 5. Therefore the other matching index is s = 9. Option A, 7, does not equal 14 - 5. Option B, 8, is also not the complementary index. Option C, 9, satisfies the symmetry rule and is correct. Option D, 10, is likewise not 14 - 5. The factorial formula confirms this: nCr = n!/[r!(n-r)!]. Replacing r by n-r merely swaps the two factors in the denominator, so the value remains unchanged. Memory cue: in nCr, the paired indices always add to n.

A committee is a group formed by selecting members, and the order in which those members are listed or chosen does not change the committee. Therefore combinations, written as ⁿCᵣ, are used for this situation. If A, B and C form a committee, the sets ABC, BCA and CAB represent the same committee. Option C is correct. A rank or an arrangement normally depends on position, so permutations are more appropriate there. A password also depends on the order of its characters and may involve repetition, so it is not an ordinary combination situation. The key governing idea is that combinations count selections where order is irrelevant, whereas permutations count ordered arrangements.

The governing concept is the symmetry property of combinations: ⁿCᵣ = ⁿCₙ₋ᵣ. It means that selecting r objects from n objects has the same count as leaving out n−r objects. Here n = 13 and r = 4, so the complementary lower index is 13 − 4 = 9. Therefore ¹³C₄ = ¹³C₉, and because the question explicitly says m ≠ 4, the required value is m = 9. Thus option B is correct. Option A gives ¹³C₈, whose complementary index is 5, not 4. Likewise, the complements of 10 and 11 are 3 and 2. The symmetry identity establishes the answer directly without expanding factorials, and the condition m ≠ 4 removes the original, obvious index.