Combinations

Correct answer: D. (792)

The coefficient is ( \binom{12}{5} ), so the answer is (792). In exams, for ( (1+x)^n ), the coefficient of (x^r) is ( \binom{n}{r} ).

Correct answer: B. (596)

Taking the number of women as (2,3,4,5) and adding gives (596). In such questions, add all valid cases separately.

Correct answer: D. (6)

By symmetry, (4=n-2), so (n=6). In exams, when ( \binom{n}{r}=\binom{n}{s} ), check (r=s) or (r+s=n).

Correct answer: A. (35)

The total segments are ( \binom{10}{2} ), and subtracting (10) sides gives (35). For polygons, first count all vertex pairs.

Correct answer: A. (56)

The no-consecutive formula gives ( \binom{10-3+1}{3}=\binom{8}{3}=56 ). In such questions, the gap method is very fast.

Correct answer: D. (103776)

The selection is ( \binom{4}{2}\binom{48}{3}=103776 ). For exactly-type questions, choose special objects first and then the remaining objects.

Correct answer: B. (210)

Subtract the impossible ( \binom{5}{3} ) from total ( \binom{12}{3} ), giving (210). A triangle needs three non-collinear points.

Correct answer: A. (63)

There are (2^6) subsets, and removing the empty selection gives (63). For at least one, remember to subtract the empty case.

Correct answer: D. (270)

The three cases of color distribution (2,1,1) add up to (270). When capacities are limited, keep the cases by color separate.

Correct answer: B. (56)

Choose the (3) up-step positions among (8) total steps, so ( \binom{8}{3}=56 ). In paths, selecting step positions is a combination idea.

Correct answer: A. (36)

After subtracting minimum values, (a+b+z=7), so ( \binom{9}{2}=36 ). First convert constraints to zero-based variables.

Correct answer: D. (21)

From total ( \binom{12}{2} ), subtract (3\binom{6}{2}) cases where a variable is (6) or more, giving (21). For upper bounds, complementary counting is useful.

Correct answer: D. (881)

Taking vowels as (2,3,4,5) gives a total of (881). For an at-least condition, add all permitted counts.

Correct answer: B. (256)

Odd- and even-sized subsets are equal in number, so the count is (2^8=256). In such questions, half the subsets have odd size.

Correct answer: A. 84

Only a set of 3 distinct digits is required, so the order of selection does not matter. Therefore, the number of sets is \(\binom{9}{3}=\frac{9\times8\times7}{3\times2\times1}=84\). Option 504 equals \(9P3\), which applies when different orders of the selected digits are counted separately. Exam tip: When you see “set” or “choose,” first check whether order matters.

Correct answer: A. (672)

From total ( \binom{12}{5} ), subtract ( \binom{10}{3} ) selections containing both particular persons, giving (672). For not-together conditions, subtract forbidden cases.

Correct answer: D. (336)

Subtract ( \binom{9}{5} ), where neither (A) nor (B) is selected, from total ( \binom{11}{5} ), giving (336). For at least one, complement is simplest.

Correct answer: B. (420)

The condition gives (2) boys and (4) girls, so ( \binom{8}{2}\binom{6}{4}=420 ). First find the distribution of numbers.

Correct answer: B. (150)

Choosing two horizontal lines from (6) and two vertical lines from (5) gives ( \binom{6}{2}\binom{5}{2}=150 ). A rectangle needs two horizontal and two vertical lines.

Correct answer: D. (60)

A parallelogram needs (2) lines from each family, so ( \binom{4}{2}\binom{5}{2}=60 ). The key idea is choosing two parallel lines from each family.

Correct answer: A. (495)

The formula gives ( \binom{15-4+1}{4}=\binom{12}{4}=495 ). For no-consecutive selections, take combinations from reduced positions.

Correct answer: B. (28)

The sum is even when (3) even numbers or (2) odd and (1) even number are chosen, giving (28). Count parity cases separately.

Correct answer: D. (5148)

From one suit there are ( \binom{13}{5} ) selections and there are (4) suits, so the count is (5148). Choose the suit and then the cards.

Correct answer: A. (1260)

Direct answer: Option A, 1,260 selections. The fixed group contains 5 objects, while the remaining outside group contains \(14-5=9\) objects. Exactly 3 objects must come from the fixed group, so choose them in \(\binom{5}{3}=10\) ways. Seven objects are chosen in all, leaving \(7-3=4\) objects to be selected from the 9 outside objects; this gives \(\binom{9}{4}=126\) ways. The multiplication rule gives \(10\times126=1260\). Option A matches. Option B, 1080, option C, 1170, and option D, 1350 do not equal this product, so they count an incorrect number of selections. “Exactly 3” excludes choices containing 0, 1, 2, 4, or 5 objects from the fixed group. The important method is to separate the required group and its complement before applying combinations.

Correct answer: A. 5

The governing concept is the symmetry of binomial coefficients: C(n,k) = C(n,n-k). In the equation C(12,r) = C(12,r+2), the lower indices r and r+2 are different, so equality occurs when they are complementary with respect to 12. Therefore, r + (r + 2) = 12. Simplifying gives 2r + 2 = 12, then 2r = 10 and r = 5. A direct verification confirms the result: C(12,5) = C(12,7), so both sides are equal. Hence option A is correct. Equating r with r+2 would be impossible, while r = 4, 6, or 7 does not make the two lower indices complementary. The obtained value also lies in the valid range for binomial coefficients.