Combinations

Correct answer: C. 10

Use the symmetry property of combinations: \(^{n}C_r=^{n}C_{n-r}\). For \(^{n}C_4=^{n}C_6\), we need \(6=n-4\). Hence, \(n=10\). Option 8 is not correct because the symmetric counterpart of \(^{8}C_4\) is \(^{8}C_4\), not \(^{8}C_6\). Exam tip: When two \(nCr\) values with different lower indices are equal, their lower indices usually add up to \(n\).

Correct answer: B. \(4\)

Use the identity \({}^nC_r={}^nC_{n-r}\). Hence, in \({}^{12}C_r={} ^{12}C_{r+4}\), the two lower indices must be complementary: \(r+4=12-r\). Thus \(2r=8\), so \(r=4\). For example, \(r=5\) would compare \({}^{12}C_5\) and \({}^{12}C_9\), which are not equal because 5 and 9 do not add to 12. Exam tip: for \({}^nC_a={}^nC_b\), usually check whether \(a=b\) or \(a+b=n\).

Correct answer: A. 495

Given \(^{n}C_2=66\), we have \(\frac{n(n-1)}{2}=66\), so \(n(n-1)=132\). Hence \(n=12\). Therefore, \(^{12}C_4=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\). Thus, 495 is correct. The value 220 is \(^{12}C_3\), not \(^{12}C_4\). Exam tip: first find \(n\) from \(^{n}C_2\), then evaluate the required combination.

Correct answer: D. (21)

(\frac{^{n}C_{5}}{^{n}C_{4}}=\frac{n-4}{5}=\frac{7}{2}) gives no integer (n). Always check validity of the ratio.

Correct answer: C. (372)

If both special students are selected, count (^{10}C_{3}); if both are not selected, count (^{10}C_{5}). Total (120+252=372).

Correct answer: C. (560)

Cases are (3W2M), (4W1M), and (5W0M). The value is (^{6}C_{3}{}^{8}C_{2}+^{6}C_{4}{}^{8}C_{1}+^{6}C_{5}=560).

Correct answer: C. 84

To form a triangle, we choose any 3 of the 9 points. Since no 3 points are collinear, every such selection forms a triangle. Therefore, the total number of triangles is
\(\binom{9}{3}=\frac{9\times8\times7}{3\times2\times1}=84\). The order of choosing the points does not matter, so combinations—not permutations—are used. Exam tip: When no three points are collinear, every choice of three points gives a triangle.

Correct answer: A. \(\binom{8}{3}\binom{7}{2}+\binom{8}{4}\binom{7}{1}+\binom{8}{5}\)

There are three mutually exclusive cases: exactly 3, 4, or 5 women. Multiply the choices of women and men in each case, then add the results. Option B counts committees with at least one woman, not at least three. Exam tip: for “at least,” list every allowed case before forming the sum.

Correct answer: C. 495

Every selection of 4 vertices of a convex 12-sided polygon determines exactly one quadrilateral. Therefore, the required number is \(\binom{12}{4}=495\). Since the order of selecting vertices does not matter, combinations—not permutations—are used. For instance, 450 equals \(\binom{10}{2}\), which does not represent choosing 4 vertices here. Exam tip: The number of quadrilaterals formed from a convex \(n\)-gon is \(\binom{n}{4}\).

Correct answer: A. 1583

There are \(\binom{13}{6}=1716\) ways to choose 6 players from all 13 players. Exclude groups containing 0 or 1 bowler. With 0 bowlers, the number of groups is \(\binom{7}{6}=7\). With 1 bowler, it is \(\binom{6}{1}\binom{7}{5}=6\times21=126\). Hence the required number is \(1716-7-126=1583\). The option 1589 results from an incorrect count of the one-bowler cases. Exam tip: for an “at least” condition, counting by complement is often quicker.

Correct answer: C. (310)

Subtract all-Indian (^{6}C_{4}=15) and all-foreign (^{5}C_{4}=5) from total (^{11}C_{4}=330). Answer (310).

Correct answer: A. (875)

Choose (3,4,5) from the first (5) and complete from the remaining (10). (^{5}C_{3}{}^{10}C_{7}+^{5}C_{4}{}^{10}C_{6}+^{5}C_{5}{}^{10}C_{5}=875).

Correct answer: D. (945)

Cases are (3,5), (4,4), and (5,3). Sum (^{6}C_{3}{}^{7}C_{5}+^{6}C_{4}{}^{7}C_{4}+^{6}C_{5}{}^{7}C_{3}=945).

Correct answer: A. (105)

First count all groups of four that can be selected from nine scientists, without considering the restriction. Then remove the groups that contain both rival scientists. A team is invalid exactly when both rivals are present; the other two members can be any two of the remaining seven scientists.

The total number is \\(\binom{9}{4}=126\\). If both rivals are included, two additional members must be chosen from the remaining seven, giving \\(\binom{7}{2}=21\\) invalid teams. Hence valid teams number \\(126-21=105\\). Therefore option A is correct. Subtracting only one team would ignore the many possible choices for the other members.

Correct answer: C. 2170

Since each colour must be represented among 5 balls, the only possible colour distributions are (3,1,1) and (2,2,1). For (3,1,1), the number of selections is \(\binom63\binom51\binom41+\binom61\binom53\binom41+\binom61\binom51\binom43=760\). For (2,2,1), it is \(\binom62\binom52\binom41+\binom62\binom51\binom42+\binom61\binom52\binom42=1410\). Hence, the total is \(760+1410=2170\). Option 1410 counts only the second distribution. Exam tip: when listing colour distributions, include every assignment of the quantities to the colours.

Correct answer: B. (16128)

Total is (^{18}C_{6}=18564), and selections with both special members are (^{16}C_{4}=1820), giving (16744). None of the options is correct.

Correct answer: C. (18144)

Choose (1) of the two special students and (5) from the remaining (18). The number is (^{2}C_{1}{}^{18}C_{5}=17136).

Correct answer: A. all (n\geq 3)

While choosing (3) from two groups of size (n), either all (3) come from one group or the split is (2,1). This proves the identity.

Correct answer: A. \(2r-2\)

Using the symmetry property of combinations, if \(^{n}C_a=^{n}C_b\) and \(a\ne b\), then \(a+b=n\). Here, \(r\ne r-2\), so \(r+(r-2)=n\). Therefore, \(n=2r-2\). Taking \(2r\) would not make \(r-2\) the complementary index of \(r\). Exam tip: use the identity \(^{n}C_r=^{n}C_{n-r}\).

Correct answer: B. (324)

Total is (^{14}C_{3}=364). Invalid selections are (^{6}C_{3}+^{4}C_{3}=24), so the answer is (340).

Correct answer: D. (40)

Total pairs are (^{10}C_{2}=45). The (^{4}C_{2}=6) pairs among collinear points give only (1) line, so (45-6+1=40).

Correct answer: A. 126

To reach the destination, 5 right moves and 4 up moves are needed, making 9 moves in all. A path is determined by choosing the positions of the 5 right moves among these 9 moves. Hence, the number of paths is \(\binom{9}{5}=126\). Option 120 equals \(5!\), but that counts arrangements of only 5 objects; here the two types of moves must be arranged among all 9 positions. Exam tip: In such grid-path questions, choose the positions of one type of move from the total moves.

Correct answer: B. (700)

First part has (^{5}C_{2}=10) paths and the second part has (^{6}C_{4}=15) paths. Total is (150).

Correct answer: C. (3876)

If no two are consecutive, use (^{n-r+1}C_{r}). Here (^{17}C_{4}=2380), so the correct answer is (2380).

Correct answer: A. (462)

By the gap method, the count is (^{15-5+1}C_{5}=^{11}C_{5}=462). In such problems identify the gap condition first.