Combinations

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

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.

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.

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

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.

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

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

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

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

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.

(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.

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

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

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

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.

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.

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.

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.

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.

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

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.

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

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

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

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