Combinations

Correct answer: B. (12264)

Total ways are (\binom{17}{6}=12376). Removing all-white (\binom{8}{6}=28) and all-black (\binom{9}{6}=84) gives (12264).

Correct answer: A. 1287

Since all prizes are identical, the order of the 5 selected candidates does not matter. Hence, use combinations: \(\binom{13}{5}=\frac{13!}{5!8!}=1287\). The value 154440 is \(^{13}P_5\), which would apply if the prizes were distinct. Exam tip: When identical objects or prizes are involved, order is not counted, so use combinations.

Correct answer: B. (406)

Total ways are (\binom{11}{5}=462) and if no special friend goes then (\binom{8}{5}=56). Hence (462-56=406).

Correct answer: C. (350)

Choose (2) from the special fruits and (3) from the remaining (7). The ways are (\binom{5}{2}\binom{7}{3}=350).

Correct answer: C. (5292)

A complete selection requires two red balls, two blue balls, and one green ball. The choices within each color are unordered, so each part is counted with a combination. Because the color groups are separate, the number of complete selections is the product of the three independent counts. This is the standard use of combinations together with the multiplication principle.

The red choices are \(\binom{7}{2}=21\), the blue choices are \(\binom{8}{2}=28\), and the green choices are \(\binom{9}{1}=9\). Therefore the total number is \(21\times28\times9=5292\). Thus option C is correct. The factors should be multiplied rather than added, since every red choice can be paired with every blue and green choice.

Correct answer: B. (40)

The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (3) from the remaining (6), so (\binom{2}{1}\binom{6}{3}=40).

Correct answer: B. (50)

Chemistry is fixed so choose the remaining (3) subjects from (8). Removing (6) selections containing both mathematics and physics leaves (50) ways.

Correct answer: B. (330)

The (2) numbers are fixed and (2) are excluded. The remaining (4) numbers are chosen from (11) in (\binom{11}{4}=330) ways.

Correct answer: B. (\binom{8}{3})

For a failed triangle (3) points are chosen from the (8) collinear points. So the number is (\binom{8}{3}).

Correct answer: A. (28)

After removing (4) broken chairs (8) good chairs remain. The ways to choose (6) are (\binom{8}{6}=28).

Correct answer: B. (420)

If both are included there are (\binom{10}{4}=210) ways and if both are excluded there are (\binom{10}{6}=210) ways. The total is (420).

Correct answer: C. (546)

You can choose (3), (4), or (5) from the last (5) questions. The total is (\binom{5}{3}\binom{7}{4}+\binom{5}{4}\binom{7}{3}+\binom{5}{5}\binom{7}{2}=546).

Correct answer: C. (8106)

The number of seniors can be (0), (1), or (2). The total is (\binom{8}{0}\binom{10}{6}+\binom{8}{1}\binom{10}{5}+\binom{8}{2}\binom{10}{4}=8106).

Correct answer: B. (660)

Choose (1) of the two fixed cards and (4) cards from the remaining (11). The ways are (\binom{2}{1}\binom{11}{4}=660).

Correct answer: C. 2048

The total number of subsets of 12 coins is \(2^{12}\). The numbers of selections of even size and odd size are equal, so the number of even-size selections is \(\frac{2^{12}}{2}=2^{11}=2048\). This includes the selection of 0 coins; excluding it would give 2047. Exam tip: for \(n\) distinct objects, the number of even-sized subsets is \(2^{n-1}\).

Correct answer: C. (1024)

The number of odd selections is (2^{11-1}=1024). Remember that even and odd selections are equal in such questions.

Correct answer: B. (126)

The captain is excluded and the vice-captain is fixed. The remaining (4) players are chosen from (9), so (\binom{9}{4}=126).

Correct answer: C. (6072)

The number of special objects can be (0), (1), or (2). The total is (\binom{5}{0}\binom{11}{6}+\binom{5}{1}\binom{11}{5}+\binom{5}{2}\binom{11}{4}=6072).

Correct answer: B. (1074)

Total triples are (\binom{20}{3}=1140) and failed triples are (\binom{8}{3}+\binom{5}{3}=66). Hence (1140-66=1074).

Correct answer: C. (15750)

The number of men can be (2), (3), or (4). The total is (\binom{10}{2}\binom{8}{4}+\binom{10}{3}\binom{8}{3}+\binom{10}{4}\binom{8}{2}=15750).

Correct answer: A. 590

This is a restricted-combination problem because five books must be selected and all three subjects must be represented. The possible subject distributions are (3,1,1), (1,3,1), (1,1,3), (2,2,1), (2,1,2), and (1,2,2). Their counts are respectively C(5,3)C(4,1)C(3,1)=120, C(5,1)C(4,3)C(3,1)=60, C(5,1)C(4,1)C(3,3)=20, C(5,2)C(4,2)C(3,1)=180, C(5,2)C(4,1)C(3,2)=120, and C(5,1)C(4,2)C(3,2)=90. Adding these mutually exclusive cases gives 120+60+20+180+120+90=590. Thus option A is correct. The restriction is essential; simply choosing any five of the twelve books would include selections missing one subject.

Correct answer: C. (2100)

The cases are (2) special and (3) special students. The total is (\binom{4}{2}\binom{10}{4}+\binom{4}{3}\binom{10}{3}=2100).

Correct answer: B. (6960)

The number of special objects can be (3), (4), (5), or (6). The total is (\sum_{r=3}^{6}\binom{6}{r}\binom{10}{7-r}=6960).

Correct answer: C. (300)

Choose (1) from (a,b), then choose the remaining (4) so that (c,d) are not both included. The correct expression is \(\binom{2}{1}\left(\binom{10}{4}-\binom{8}{2}\right)=364\).

Correct answer: A. (122760)

Direct answer: Option A, 122,760 ways. The multiples of 4 from 1 through 40 are 4, 8, ..., 40. Since \(40\div4=10\), there are 10 such numbers. The remaining \(40-10=30\) numbers are not multiples of 4. Exactly 2 multiples must be selected, so their choices are \(\binom{10}{2}=45\). The other 3 numbers must come from the 30 non-multiples, giving \(\binom{30}{3}=4060\). Thus the total is \(45\times4060=122760\). Option A matches this result. Options B, 130,500, C, 136,080, and D, 182,700, do not equal the required product and therefore count the wrong number of selections. The word “exactly” is essential: selecting more or fewer than two multiples is not allowed. Memory cue: count the special category first, then choose the remaining items from its complement.