Combinations

Correct answer: A. (\binom{9}{4})

By Pascal's identity (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the answer is (\binom{9}{4}).

Correct answer: A. 84

Using Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\), we get \(\binom{10}{4}-\binom{9}{4}=\binom{9}{3}=84\). Hence, the correct answer is 84. The value 126 is \(\binom{9}{4}\), not the required difference. Exam tip: Recognising Pascal’s identity avoids lengthy calculation in such questions.

Correct answer: C. (15)

The element (1) is already chosen so the remaining (2) elements are chosen from (6). The number of ways is (\binom{6}{2}=15).

Correct answer: B. (55)

The direct answer is option B, 55. There are 8 elements in the set, and we choose 4, so the unrestricted number is C4 = 8!/(4!4!) = 70. The phrase “do not contain both 2 and 3 together” means a subset is forbidden only when it contains both 2 and 3; it may contain neither or exactly one of them. Count the forbidden subsets: if 2 and 3 are already included, choose the remaining 2 elements from the other 6, giving C2 = 15. Hence allowed subsets = 70 − 15 = 55. Option A, 50, subtracts too many or uses a wrong combination. Option B, 55, is correct. Option C, 60, does not remove all forbidden subsets. Option D, 70, counts every subset and therefore includes the 15 invalid ones. A common language mistake is to read the condition as “exclude both separately”; that would be a different question. Memory cue: “not both” means subtract only the cases containing the pair together.

Correct answer: C. 20

The even numbers from 1 to 12 are 2, 4, 6, 8, 10 and 12, so there are 6 eligible numbers. The number of ways to choose 3 of them is \(\binom{6}{3}=20\). Therefore, the correct answer is 20. A value such as 24 may result from counting arrangements, but here only selection is required and order does not matter. Exam tip: when a question asks for selection without considering order, use combinations \(\binom{n}{r}\).

Correct answer: A. 196

From 1 to 15, there are 8 odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, and 7 even numbers. The number of ways to choose 2 of the 8 odd numbers is \(\binom{8}{2}=28\), and the number of ways to choose 1 of the 7 even numbers is \(\binom{7}{1}=7\). Therefore, the total number of ways is \(\binom{8}{2}\binom{7}{1}=28\times7=196\). The value 224 can result from an incorrect count. Exam tip: multiply combinations when selections are made independently from separate groups.

Correct answer: A. (252)

After removing (3) special students (10) students remain. Hence there are (\binom{10}{5}=252) ways.

Correct answer: B. (1554)

Total ways are (\binom{14}{5}=2002). Remove (0) teacher (\binom{8}{5}=56) and (1) teacher (\binom{6}{1}\binom{8}{4}=420). The answer is (1526).

Correct answer: D. (320)

Total ways are (\binom{11}{4}=330). Removing all-white (\binom{5}{4}=5) and all-black (\binom{6}{4}=15) gives (310).

Correct answer: A. 120

Since the prizes are identical, the order of the 3 selected candidates does not matter. Therefore, use combinations: \(\binom{10}{3}=\frac{10\times9\times8}{3\times2\times1}=120\). Hence, the correct answer is 120. The option \(360\) is a permutation-based distractor, but order is not counted when prizes are identical. Exam tip: For identical prizes or objects, usually think of combinations.

Correct answer: B. (55)

The direct answer is option B, 55. We select 4 friends from 8, and at least one of two particular friends must be selected. Use the complement. First count all four-friend groups: C4 = 8!/(4!4!) = 70. Now count the groups in which neither of the two particular friends goes. Then all 4 members must come from the other 6 friends, giving C4 = 6!/(4!2!) = 15. Subtracting gives 70 − 15 = 55. Option A, 45, subtracts an incorrect number. Option B, 55, is correct because it excludes exactly the groups missing both required candidates. Option C, 60, does not remove all such groups. Option D, 70, is the unrestricted total and includes groups where both particular friends are absent. “At least one” allows exactly one or both, so we must not exclude groups containing both. Memory cue: at least one of a pair = total groups minus groups containing neither.

Correct answer: B. (60)

The direct answer is option B, 60. Exactly one of the three special fruits must be selected. First choose that one special fruit in C1 = 3 ways. After this, we need three more fruits, and they must come from the six non-special fruits, because choosing another special fruit would violate “exactly one.” Those three can be chosen in C3 = 20 ways. By the multiplication principle, total ways are 3×20 = 60. Option A, 45, is too small and does not represent the required product. Option B, 60, is correct. Option C, 75, uses an incorrect combination count. Option D, 90, overcounts, usually by failing to enforce that the other three fruits are non-special. “Exactly one” is stricter than “at least one”: no second special fruit may be included. Memory cue: exactly one from a special group means choose one special item, then choose all remaining items outside that group.

Correct answer: C. 120

There are 4 choices for the red ball, 5 for the blue ball, and 6 for the green ball. By the multiplication principle, the total number of ways is \(\binom{4}{1}\binom{5}{1}\binom{6}{1}=4\times5\times6=120\). The value 60 can result from multiplying choices for only two colours, so it does not represent the complete selection. Exam tip: When one item is chosen from each independent group, multiply the numbers of choices.

Correct answer: B. (10)

The letter (a) is fixed and (b) is excluded so choose the remaining (2) letters from (5). The ways are (\binom{5}{2}=10).

Correct answer: C. (16)

Total ways are (\binom{6}{3}=20) and if both special subjects are included the third subject is chosen in (4) ways. Hence (20-4=16).

Correct answer: B. (28)

The (2) numbers are fixed and (1) is excluded. The remaining (2) numbers are chosen from (8) in (\binom{8}{2}=28) ways.

Correct answer: B. (\binom{5}{3})

For a failed triangle (3) points are chosen from the (5) collinear points. So the number is (\binom{5}{3}).

Correct answer: A. (5)

After removing (2) broken chairs (5) good chairs remain. The ways to choose (4) are (\binom{5}{4}=5).

Correct answer: C. (70)

If both are included there are (\binom{7}{2}=21) ways and if both are excluded there are (\binom{7}{4}=35) ways. The total is (56).

Correct answer: B. (31)

The cases are choosing (2) or (3) from the last (3). The total is (\binom{3}{2}\binom{5}{3}+\binom{3}{3}\binom{5}{2}=40).

Correct answer: D. (350)

The number of seniors can be (0) or (1). The total is (\binom{5}{0}\binom{7}{4}+\binom{5}{1}\binom{7}{3}=210).

Correct answer: A. (112)

Choose (1) of the two fixed cards and (3) cards from the remaining (8). The ways are (\binom{2}{1}\binom{8}{3}=112).

Correct answer: B. (32)

An even selection means (0), (2), (4), or (6) coins. The total is (\binom{6}{0}+\binom{6}{2}+\binom{6}{4}+\binom{6}{6}=32).

Correct answer: C. 64

The selections with an odd number of toys are \(\binom71+\binom73+\binom75+\binom77=7+35+21+1=64\). In general, for \(n\) distinct objects, the numbers of even-sized and odd-sized selections are equal, each being \(2^{n-1}\). Option 127 is \(2^7-1\), the number of all non-empty selections, not only odd-sized selections. Exam tip: For even/odd selection questions, use \(2^{n-1}\) directly.

Correct answer: A. 35

Since the captain is already fixed and must not be included in the selection, only 7 players are available. The number of ways to choose 3 from these 7 players is \(\binom{7}{3}=\frac{7\times6\times5}{3\times2\times1}=35\). Therefore, the correct answer is 35. \(\binom{8}{3}=56\) would apply only if the captain were also available for selection. Exam tip: First subtract anyone who is excluded from the selection.