Combinations

Either (0) or (1) special object can be chosen. The total is (\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756).

The governing concept is combinations because a matchup is determined by selecting two distinct teams, and the order of selection does not matter. The number of unordered pairs among 16 teams is C(16,2) = 16 × 15 ÷ 2 = 120. Every pair plays 3 matches, so multiply the number of pairings by 3: 120 × 3 = 360. Thus option C is correct. If the three matches were ignored, the result would be 120, not any listed answer. Counting ordered pairs would give 240 and would count each pair twice; multiplying that by 3 would overcount the schedule. Options A, B, and D therefore do not satisfy the stated conditions.

The governing concept is the multiplication principle applied to independent combinations. From 1 through 36, the multiples of 4 are 4, 8, 12, ..., 36, so their number is 36/4 = 9. Consequently, the number of non-multiples of 4 is 36 - 9 = 27. The required selection consists of 3 numbers from the first group and 2 from the second group. Since order does not matter within either group, the count is C(9,3) × C(27,2). Now C(9,3) = 84 and C(27,2) = 351, giving 84 × 351 = 29,484. Therefore option B is correct. The other options can result from using incorrect group sizes, using an inappropriate permutation count, or failing to multiply the two independent choices.

The governing concept is multiplication of independent combinations formed from disjoint subsets. The integers from 1 to 30 contain 15 odd numbers and 15 even numbers. The number of ways to choose 3 odd numbers is C(15,3) = 15×14×13/(3×2×1) = 455. Independently, the number of ways to choose 3 even numbers is also C(15,3) = 455. Every valid odd-number selection can be paired with every valid even-number selection, so the total is C(15,3) × C(15,3) = 455 × 455 = 207,025. Thus option C is correct. Choosing any six numbers by C(30,6) would ignore the parity condition, while the other listed values arise from incorrect multiplication or incomplete counting.

The governing concept is the product rule for independent combinations. The red selection can be made in C(8,2) = 8×7/2 = 28 ways. The blue selection can be made in C(9,2) = 9×8/2 = 36 ways, and the green selection can be made in C(10,2) = 10×9/2 = 45 ways. The choices from the three colours are independent, so each red pair can be combined with every blue pair and every green pair. Hence the total number of selections is 28 × 36 × 45 = 45,360. Therefore option C is correct. Permutations are inappropriate because the order in which balls are selected does not create a new selection. Selecting six balls without enforcing two of each colour would answer a different question.

The governing concept is the equal division of all subsets into even-cardinality and odd-cardinality subsets. Each of the 13 distinct toys has two possibilities: selected or not selected. Therefore, the total number of subsets is 2^13 = 8192, including the empty subset. To see why the odd and even counts are equal, pair every subset with the subset obtained by toggling one fixed toy: if the original subset has even size, the paired subset has odd size, and vice versa. Thus exactly half of all subsets have odd size. The required number is 2^13/2 = 2^12 = 4096, so option C is correct. Option D counts all subsets, and option B excludes only the empty set but still includes even-sized nonempty subsets.

The governing concept is the multiplication principle combined with combinations. Because the selection must contain exactly 3 seniors and exactly 1 junior, choose these groups separately. The number of ways to choose 3 seniors from 6 is C(6,3) = 6×5×4/(3×2×1) = 20. The number of ways to choose 1 junior from 4 is C(4,1) = 4. Every senior selection can be paired with every junior selection, so the total is 20×4 = 80. Therefore, option B is correct. Adding the groups would be wrong because both choices are required, not alternatives. Permutations are also unnecessary because the final group has no order; only its members matter.

This is a combinations problem because only the membership of the group matters; the order in which players are chosen is irrelevant. Without the restriction, the number of groups is C(10,4) = 10!/(4!6!) = 210. To count the invalid groups, include both the captain and the vice-captain first. Then the remaining 2 members must be selected from the other 8 players, giving C(8,2) = 28 invalid groups. Therefore, valid groups = 210 - 28 = 182. Hence option C is correct. Options A, B, and D do not result from correctly subtracting the groups containing both restricted players.

Since a selection is an unordered collection, combinations are used. First count all ways to choose 4 books from 7: C(7,4) = 35. Now count the forbidden selections containing both specified books. Once those two books are included, the remaining 2 books must be chosen from the other 5 books, so the number of forbidden selections is C(5,2) = 10. Subtracting these from the total gives 35 - 10 = 25 valid selections. Therefore option B is correct. Option D is the unrestricted total, while the other values do not correctly remove the selections containing both specified books.

A diagonal joins two non-adjacent vertices. One method is to choose any two of the 10 vertices, giving C(10,2) = 10 × 9 / 2 = 45 pairs. These pairs include the 10 sides, so the number of diagonals is 45 - 10 = 35. Equivalently, from each vertex we can join to 10 - 3 = 7 non-adjacent vertices, giving 10 × 7 = 70, but this counts every diagonal twice, so 70 / 2 = 35. The standard formula n(n - 3)/2 gives the same result: 10 × 7 / 2 = 35. Hence option B is correct; 45 counts sides as well as diagonals.

The phrase exactly 2 physics books fixes the composition of the selection. First choose 2 physics books from the 4 available: C(4,2) = 4×3/2 = 6. Since five books are required altogether, the remaining 3 must be mathematics books. These can be chosen from the 5 mathematics books in C(5,3) = 5×4×3/(3×2×1) = 10 ways. The two choices are independent, so the multiplication principle gives C(4,2)C(5,3) = 6×10 = 60. Therefore option B is correct. Choosing any five from all nine books would not enforce the condition, because it would also count selections containing zero, one, three, or four physics books. The exact-subject restriction must be applied before multiplying.

Use the complement principle because the condition says “at least one.” First count all selections of 5 questions from 9: C(9,5) = 126. Now count the invalid selections, in which neither question 1 nor question 2 is chosen. All 5 questions must then come from the remaining 7, giving C(7,5) = 21. Subtracting the invalid cases from the total gives 126 - 21 = 105. Therefore option A is correct. This approach automatically includes selections containing question 1, question 2, or both, without double-counting. The other options result from subtracting the wrong complement or from failing to impose the “at least one” condition.

Apply Pascal’s identity, C(n,r) + C(n,r + 1) = C(n + 1,r + 1). Here n = 15 and r = 7, so C(15,7) + C(15,8) = C(16,8). Therefore option B is correct. The symmetry property also tells us that C(15,8) = C(15,7), but that observation alone does not identify the requested single binomial coefficient as directly as Pascal’s identity does. Option A has the wrong lower index. Option C combines unrelated parameters and is not the result of the identity. Option D includes only one of the two terms, not their sum. The adjacent lower indices 7 and 8 are exactly the pattern required by Pascal’s rule.

There are 6 + 5 = 11 distinct figures in total. If there were no restriction, the number of ways to select any 2 figures would be C(11,2) = 11 × 10 / 2 = 55. The condition requires at least one line, so the only invalid selections are those containing two circles and no line. The number of such selections is C(5,2) = 5 × 4 / 2 = 10. By the complement principle, valid selections = 55 - 10 = 45. Thus option A is correct. This count includes line-line and line-circle pairs, while excluding only circle-circle pairs. The value 55 ignores the stated restriction and therefore is not valid.

The governing concept is the ratio of adjacent combinations. Using factorial definitions, (nC5)/(nC4) = [n!/(5!(n-5)!)] ÷ [n!/(4!(n-4)!)] = (n - 4)/5. The question gives nC5 : nC4 = 11 : 5, so the ratio is 11/5. Equate the expressions: (n - 4)/5 = 11/5. Multiplying by 5 gives n - 4 = 11, and adding 4 yields n = 15. Therefore option B is correct. Direct expansion is unnecessary and less efficient. Checking the distractors, n = 14 gives 10/5 = 2, n = 16 gives 12/5, and n = 17 gives 13/5, so none produces 11/5. The required combination conditions are also satisfied by n = 15.

Because one particular member must be present, that member is already fixed in the committee. Only the remaining two positions need to be filled, and they can be selected from the other 17 members. Since a committee has no order, use combinations rather than permutations: 17C2 = 17 × 16 / (2 × 1) = 136. Thus the number of possible committees is 136, making option A correct. A common error is to calculate 18C3, which counts committees without enforcing the fixed-member condition and gives 816. Another error is to arrange the selected members, which would count the same committee several times. Fixing the required member first removes that ambiguity.

The committee must contain exactly 2 girls, so first choose 2 girls from the 5 available: ⁵C₂ ways. Its total size is 5, so the remaining 3 members must be boys; these can be chosen from 6 boys in ⁶C₃ ways. By the multiplication principle, the required expression is ⁵C₂ × ⁶C₃, making option A correct. Because a committee is an unordered group, combinations rather than permutations must be used. ¹¹C₅ ignores the exact requirement and allows any number of girls. ⁵P₂ × ⁶P₃ treats committee membership as ordered, and option D selects 3 girls and 2 boys, which is the opposite of the stated condition.

The governing idea is combinations, because a committee is selected without regard to order. Exactly 3 of the 7 boys must be chosen, which can be done in ⁷C₃ ways. The committee has 5 members, so the remaining 2 members must be selected from the 6 girls in ⁶C₂ ways. By the multiplication principle, the required expression is ⁷C₃ × ⁶C₂, option A. Option B allows any composition of the committee and does not enforce exactly 3 boys. Option C counts ordered selections, which is inappropriate for a committee, while option D selects the wrong numbers of boys and girls.