Combinations

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 \(n\).

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 \(a+b=n\).

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.

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

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

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).

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.

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.

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 \(\binom{n}{4}\).

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.

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

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).

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).

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.

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.

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.

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

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.

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 \(^{n}C_r=^{n}C_{n-r}\).

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

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).

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.

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

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

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