Combinations

Correct answer: B. (7)

By symmetry, (r+(r-3)=11), so (r=7). In ( \binom{n}{a}=\binom{n}{b} ), often (a+b=n) is useful.

Correct answer: D. (1120)

First choose (4) couples and then (1) person from each, giving ( \binom{8}{4}2^4=1120 ). Choosing only one from a couple gives (2) choices.

Correct answer: C. (480)

Choose the couple ( \binom{6}{1} ), then (3) other couples ( \binom{5}{3} ), and one person from each (2^3), giving (480). After fixing exactly one couple, choose only one person from each remaining couple.

Correct answer: A. (140)

The cases of exactly (2) special and exactly (3) special persons add to (140). For at least (2), add both valid cases.

Correct answer: B. (84)

This is the number of nonnegative solutions of (x_1+x_2+x_3+x_4=6), so ( \binom{9}{3}=84 ). For unlimited identical types, use stars and bars.

Correct answer: C. (28)

After giving (2) pens to each student, (6) pens remain, so ( \binom{8}{2}=28 ). Subtract the minimum condition and then count nonnegative solutions.

Correct answer: D. (455)

After subtracting minimums (5,1,1,1), (12) remains, so ( \binom{15}{3}=455 ). Subtract unequal minimum conditions first.

Correct answer: C. (875)

Subtract ( \binom{15}{3} ) cases with (x_1\ge6) from total ( \binom{21}{3} ), giving (875). For upper limits, subtract forbidden solutions.

Correct answer: A. (111)

Subtract ( \binom{6}{2} ) subsets containing all three special elements from total ( \binom{9}{5} ), giving (111). In not-type questions, subtract forbidden selections.

Correct answer: C. (60)

The direct answer is option C, 60. There are 3 special elements and therefore \(9-3=6\) ordinary elements. To form a 5-element subset containing exactly 2 special elements, choose 2 of the 3 special elements and then choose the remaining 3 elements from the 6 ordinary elements: \(\binom{3}{2}\binom{6}{3}=3\times20=60\). Option A, 45, is not this product. Option B, 54, also does not count the required choices. Option C is correct because it enforces exactly 2 special elements. Option D, 72, overcounts. Choosing 3 special elements would violate “exactly 2,” while choosing fewer than 2 would not meet the condition. Memory cue: separate the required categories, choose within each, then multiply.

Correct answer: B. (256)

(a) is fixed as selected and (b) is fixed as not selected, so the remaining (8) elements are free, giving (2^8=256). Fix compulsory states first.

Correct answer: C. (456)

Taking the number of special elements as (2,3,4) gives (456). With a small special group, directly add valid cases.

Correct answer: C. (70)

The product is odd only when all four numbers are odd, so ( \binom{8}{4}=70 ). For product parity, check every factor.

Correct answer: B. (570)

For an odd sum, choose (1) odd and (2) even or (3) odd numbers, giving (570). Making a parity case table reduces errors.

Correct answer: C. (255)

For an even sum, the number of odd selected numbers can be (0,2,4), giving (255). Focus on how many odd elements are selected, not only on total selection size.

Correct answer: C. (4200)

The first committee can be formed in ( \binom{10}{3} ) ways and the second in ( \binom{7}{4} ) ways, giving (4200). For named committees, the stages matter.

Correct answer: C. (792)

If both are selected, count ( \binom{11}{4} ); if neither is selected, count ( \binom{11}{6} ), giving (792). Split either both or neither into two cases.

Correct answer: A. (140)

There are (2) ways to choose the special person and then (4) from the other (8), so (2\binom{8}{4}=140). Exactly one means two symmetric cases.

Correct answer: C. (210)

A quadrilateral needs (2) points from each line, so ( \binom{7}{2}\binom{5}{2}=210 ). Taking (3) points from one line cannot form a quadrilateral.

Correct answer: C. (21)

Using the limits (M\le1), (I\le4), (S\le4), and (P\le2), the total selections are (21). For repeated letters, count bounded integer solutions.

Correct answer: B. (3)

The valid selections are (A^3N^2), (BA^2N^2), and (BA^3N), so the total is (3). In small repetition questions, list valid distributions directly.

Correct answer: D. (495)

The product is ( (1+x)^{12} ), so the coefficient is ( \binom{12}{8}=495 ). In binomial products, combine powers first.

Correct answer: C. (1001)

By Vandermonde's identity, the sum is ( \binom{14}{4}=1001 ). In such sums, view it as total selection after adding upper counts.

Correct answer: C. (210)

Since ( \binom{10}{2}=45 ), (n=10), so ( \binom{10}{4}=210 ). First find (n), then evaluate the required combination.

Correct answer: B. (14)

The ratio is (\frac{^{n}C_{3}}{^{n}C_{2}}=\frac{n-2}{3}), so (n=14). In exams write the ratio first.