Combinations

Total pairs are (^{12}C_{2}). The (^{5}C_{2}) pairs from collinear points give only (1) line, so (66-10+1=57).

The ways to choose two vertices are (^{18}C_{2}). Subtracting (18) sides gives (135) diagonals.

Any selection of 4 points on a circle forms a quadrilateral because no three of the selected points are collinear. Therefore, the number of quadrilaterals is \,\(\binom{11}{4}=330\). Option 286 equals \,\(\binom{13}{3}\), so it does not represent choosing 4 points from 11. Exam tip: For polygons formed by points on a circle, use combinations by choosing the required number of vertices.

In a convex polygon, an interior intersection of two diagonals is determined by four distinct vertices: the endpoints of one diagonal and the endpoints of the other. For any chosen set of four vertices, the two diagonals joining opposite vertices cross once inside the polygon. The condition that no three diagonals meet at one interior point ensures that different groups of four vertices do not represent the same intersection.

Thus we choose four vertices from ten, giving \(\binom{10}{4}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\). Hence option B is correct. Counting pairs of diagonals directly would require care because some pairs may share endpoints; the four-vertex method avoids that issue and counts each interior crossing exactly once.

The circular non-adjacent selection formula is (\frac{n}{n-r},{}^{n-r}C_{r}). Here (\frac{12}{8}\times{}^{8}C_{4}=105).

A shortest path has 11 moves in total: 7 right moves and 4 up moves. Each path is determined by choosing the 7 positions for the right moves among these 11 positions. Therefore, the number of paths is \(\binom{11}{7}=\binom{11}{4}=330\). The value \(252=\binom{10}{5}\), so it does not apply here. Exam tip: In grid-path questions, choose the positions of one type of move among all moves.

Total paths are (^{11}C_{6}=462). Subtract paths through that point (^{4}C_{2}\times{}^{7}C_{4}=210), giving (252).

Let the selected numbers be \(x_1<x_2<\cdots<x_5\). Since no two are consecutive, \(x_{i+1}\ge x_i+2\). Subtracting the number of earlier selections from each later selected position converts the problem into choosing 5 positions from 21 positions. Hence, the number of selections is \(\binom{25-5+1}{5}=\binom{21}{5}=20349\). Using \(\binom{25}{5}\) would be incorrect because it also counts selections containing consecutive numbers. Exam tip: for choosing \(r\) non-consecutive objects from \(n\) objects, use \(\binom{n-r+1}{r}\).

For choosing \(r\) non-adjacent points from \(n\) points arranged on a circle, the count is \(\frac{n}{n-r}\binom{n-r}{r}\). Thus, \(\frac{10}{10-3}\binom{7}{3}=\frac{10}{7}\times35=50\). Hence, option C is correct. A value such as 42 can result from treating the arrangement as linear or from incomplete counting, but on a circle the first and last points are also adjacent. Exam tip: always account for the adjacency of the first and last positions in circular arrangements.

With at least 3 women, the possible cases are 3, 4, or 5 women. They require 2, 1, or 0 men respectively, so all three product terms are added. Exam tip: check that each case totals the committee size of 5.

This is a positive-integer solution problem. First assign 1 to each of the five variables; the remaining sum is 15. The number of non-negative integer solutions is \\(\binom{15+5-1}{5-1}=\binom{19}{4}=3876\\). Hence, 3876 is correct. Using \\(\binom{15}{4}\\) would be incorrect because the initial 1 assigned to every variable must be accounted for. Exam tip: for positive integer solutions of a sum of n into r parts, use \\(\binom{n-1}{r-1}\\).

Without the upper restrictions, the number of non-negative solutions of \(x_1+x_2+x_3+x_4=13\) is found by stars and bars: \(\binom{13+4-1}{4-1}=\binom{16}{3}=560\). We must remove solutions in which at least one variable is 6 or more. If a particular variable is at least 6, subtract 6 from it; the remaining sum is 7, giving \(\binom{10}{3}=120\) solutions for each variable.

There are four such single-variable cases, so subtract \(4\times120=480\). Two variables can both be at least 6 only when the remaining sum is 1, giving \(\binom{4}{3}=4\) solutions for each of the \(\binom{4}{2}=6\) pairs, or 24 total. Three variables cannot reach 6 because their minimum sum would be 18. Thus \(560-480+24=104\), so option B is correct.

Without the restriction on \(x\), the number of non-negative solutions of \(x+y+z=10\) is found by stars and bars: \(\binom{10+3-1}{3-1}=\binom{12}{2}=66\). We must remove the solutions where \(x\geq4\), because the required condition is \(x\leq3\). Let \(x'=x-4\), so \(x'\) is non-negative.

The changed equation is \(x'+y+z=6\), whose number of solutions is \(\binom{8}{2}=28\). Hence the desired number is \(66-28=38\). This counts every non-negative solution exactly once and removes precisely those with an excessive value of \(x\). Therefore option C is correct.

First choose one of (4) suits and then choose (5) cards from (13). The number is (4\times{}^{13}C_{5}=5148).

Choose (2) hearts in (^{13}C_{2}) ways and the other (3) cards in (^{39}C_{3}) ways. Multiplication gives (712842).

Use the complement method for hands containing at least one ace. The total number of 5-card hands is \(\binom{52}{5}=2598960\). Hands with no ace are selected from the remaining 48 cards, giving \(\binom{48}{5}=1712304\). Therefore, the required number is \(\binom{52}{5}-\binom{48}{5}=886656\). Option A counts hands with exactly one ace only, whereas the question also includes hands with two, three, or four aces. Exam tip: for “at least one” conditions, counting the complement is often quickest.

Choose (3) queens from (4) and the other (3) cards from (48) non-queen cards. The number is (^{4}C_{3}\times{}^{48}C_{3}=69184).

For an even sum, the number of odd digits must be (0), (2), or (4). The count is (^{5}C_{0}{}^{4}C_{4}+^{5}C_{2}{}^{4}C_{2}+^{5}C_{4}{}^{4}C_{0}=66).

From 1 through 15, the multiples of 3 are 3, 6, 9, 12, and 15, so there are 5 multiples. The remaining 10 numbers are not multiples of 3. We need exactly two multiples and exactly two non-multiples in a selection of four. Choose the two multiples in \\(^{5}C_2=10\\) ways and the two non-multiples in \\(^{10}C_2=45\\) ways. By multiplication, the total is \\(^{5}C_2\\times{}^{10}C_2=10\\times45=450\\).

The word exactly is important: selecting three or four multiples would not satisfy the condition. We independently choose the two required groups, and every resulting group of four is counted once. Therefore option B, 450, is correct.

For a rectangle, choose (2) horizontal lines from (6) and (2) vertical lines from (7). The count is (^{6}C_{2}\times{}^{7}C_{2}=315).

First choose (5) students in (^{10}C_{5}) ways. Since the two groups are unnamed, divide by (2) to get (126).

First choose (4) people in (^{8}C_{4}) ways and split them into two pairs in (3) ways. Total (70\times3=210).

Choose (4) couples from (6) and then choose (1) person from each selected couple. The number is (^{6}C_{4}\times2^{4}=240).

The number of identical candies can be from (0) to (4). Sum (^{8}C_{4}+^{8}C_{3}+^{8}C_{2}+^{8}C_{1}+1=163).

The direct answer is option C, 246. The 4 magazines are identical, so choosing 2 magazines is only one magazine-choice pattern, not \(^{4}C_2\). The 9 books are distinct, so their choices are counted with combinations. At least 2 magazines means the number of magazines can be 2, 3, or 4. If 2 magazines are selected, 4 books must be selected: \(^{9}C_4=126\). If 3 magazines are selected, 3 books are selected: \(^{9}C_3=84\). If 4 magazines are selected, 2 books are selected: \(^{9}C_2=36\). Adding disjoint cases gives \(126+84+36=246\). Option A, 210, misses or miscounts a case. Option B, 228, is not the correct sum. Option C, 246, includes exactly all three allowed magazine cases. Option D, 264, overcounts. The key lesson is that identical objects do not create separate choices by identity, while distinct books do.