Combinations

Correct answer: A. (225)

Cases are (2) special and (3) others, or (3) special and (2) others. The value is (^{3}C_{2}{}^{9}C_{3}+^{3}C_{3}{}^{9}C_{2}=288).

Correct answer: A. (917)

Cases are (4T2S), (5T1S), and (6T0S). Sum is (^{7}C_{4}{}^{8}C_{2}+^{7}C_{5}{}^{8}C_{1}+^{7}C_{6}=1155).

Correct answer: D. (385)

We can take (0) to (4) identical objects. The total is (^{10}C_{4}+^{10}C_{3}+^{10}C_{2}+^{10}C_{1}+1=386).

Correct answer: B. (99)

The number of identical candies can be (0) to (4). Sum (^{7}C_{4}+^{7}C_{3}+^{7}C_{2}+^{7}C_{1}+1=99).

Correct answer: A. (792)

Identical prizes with at most one per student means choosing (5) students. The number is (^{12}C_{5}=792).

Correct answer: B. 28

Each handshake is formed by choosing a pair of people from the 8 people, and the order of the two people does not matter. Therefore, the number of handshakes is \(\binom{8}{2}=\frac{8\times7}{2}=28\). Using \(8\times7=56\) would count every handshake twice, so we divide by 2. Exam tip: for “every pair exactly once” questions, usually use \(\binom{n}{2}\).

Correct answer: C. 5

If n people are present, the number of handshakes when each distinct pair shakes hands once is \(\binom{n}{2}\). Here, \(\binom{n}{2}=435\), so \(\frac{n(n-1)}{2}=435\). Thus, \(n(n-1)=870\), giving \(n=30\). Therefore, the number absent is \(35-30=5\). Option 6 would mean that only 29 people attended, but then the number of handshakes would be \(\binom{29}{2}=406\). Exam tip: use \(\binom{n}{2}\) for handshake questions when every pair meets once.

Correct answer: A. \(\binom{n}{r}=\binom{n}{n-r}\)

Choosing \(r\) objects from \(n\) is equivalent to deciding which \(n-r\) objects are left out; hence \(\binom{n}{r}=\binom{n}{n-r}\). Option D is Pascal’s identity, not symmetry. Exam tip: use complementary selection to spot this relation quickly.

Correct answer: B. Pascal's rule

Pascal's rule is \(\binom{n}{r-1}+\binom{n}{r}=\binom{n+1}{r}\). On putting \(r=3\), we get \(\binom{n}{2}+\binom{n}{3}=\binom{n+1}{3}\). Hence, the correct answer is Pascal's rule. The multiplication rule is used to multiply the numbers of choices in successive independent steps, so it does not describe this additive identity. Exam tip: A sum of two consecutive binomial coefficients producing one coefficient with upper index increased by 1 indicates Pascal's rule.

Correct answer: A. 330

The number of ways to choose 3 objects is \(\binom{10}{3}=120\), and the number of ways to choose 4 objects is \(\binom{10}{4}=210\). A selection cannot contain both exactly 3 and exactly 4 objects, so these cases are mutually exclusive and their counts are added: \(120+210=330\). Therefore, the correct answer is 330. A value such as 340 is incorrect because the required combination counts add exactly to 330. Exam tip: For mutually exclusive cases joined by “or,” add the number of ways in the separate cases.

Correct answer: A. 792

Exclude the fixed student first. We then select 5 students from the remaining 12, so the number of selections is \(\binom{12}{5}=792\). Option 924 equals \(\binom{12}{6}\), so it does not apply here. Exam tip: for a person who must not be selected, remove that person and form the combination from the remaining group.

Correct answer: C. (455)

The direct answer is option C, 455 selections. Since balls are distinct, a selection means choosing objects without caring about order. There are 12 balls altogether, so the total number of ways to choose 4 is \(^{12}C_4=495\). The condition says the selected balls must not all have one colour. The unwanted cases are: four red balls, counted by \(^{5}C_4=5\), and four black balls, counted by \(^{7}C_4=35\). These two cases cannot overlap. Therefore valid selections are \(495-5-35=455\). Option A, 410, subtracts too many or uses an incorrect count. Option B, 430, is also not the result of the required subtraction. Option C, 455, correctly removes both all-red and all-black selections. Option D, 475, does not remove the two forbidden groups correctly. Remember: for “not all of one type”, count everything and subtract every completely single-type case.

Correct answer: C. (64)

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

Correct answer: B. (110)

There are (6) odd and (6) even numbers. For odd sum choose (1) or (3) odd numbers, giving (^{6}C_{1}{}^{6}C_{2}+^{6}C_{3}=110).

Correct answer: A. (126)

The number of special players can be (0,1,2), with (7) ordinary players available. Sum (^{4}C_{0}{}^{7}C_{7}+^{4}C_{1}{}^{7}C_{6}+^{4}C_{2}{}^{7}C_{5}=155).

Correct answer: B. \(n=2r+1\)

Given \(^{n}C_r=^{n}C_{r+1}\). Using the ratio of consecutive combinations, \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\). For the two combinations to be equal, this ratio must be 1. Hence \(n-r=r+1\), so \(n=2r+1\). The relation \(n=2r\) does not generally make two consecutive combinations equal. Exam tip: use the symmetry \(^{n}C_r=^{n}C_{n-r}\) to write \(r+1=n-r\) directly.

Correct answer: C. 455

Given \(^{n}C_2=105\), we have \(\frac{n(n-1)}{2}=105\). Thus, \(n(n-1)=210\), which gives \(n=15\). Therefore, \(^{15}C_3=\frac{15\times14\times13}{3\times2\times1}=455\). Hence, 455 is correct. The nearby distractor 364 is \(^{14}C_3\), not \(^{15}C_3\). Exam tip: first determine \(n\) from \(^{n}C_2\), then calculate the required combination.

Correct answer: A. 4

Use the identity \(^{n}C_r=^{n}C_{n-r}\). Here, \(r\) and \(2r+4\) must be complementary indices, so \(r+(2r+4)=16\). Thus, \(3r=12\), giving \(r=4\). The direct-equality case gives \(r=-4\), which is not a valid index for the combination here. Exam tip: For \(^{n}C_a=^{n}C_b\), check both cases: \(a=b\) and \(a+b=n\).

Correct answer: D. 3432

Using Pascal’s identity, \(^{n}C_r+{}^{n}C_{r+1}=^{n+1}C_{r+1}\). Therefore, \(^{13}C_6+{}^{13}C_7=^{14}C_7\). Since \(^{14}C_7=3432\), option D is correct. The value 1716 is only \(^{13}C_6\), not the required sum. Exam tip: When two consecutive combinations have the same upper index, apply Pascal’s identity.

Correct answer: C. (350)

The possible cases are (2G3B) and (3G2B). The sum is (^{5}C_{2}\times{}^{6}C_{3}+^{5}C_{3}\times{}^{6}C_{2}=350).

Correct answer: D. (5880)

Choose (4) teachers using (^{10}C_{4}) and (2) students using (^{8}C_{2}). Multiplication gives (5880).

Correct answer: B. (378)

Choose (1) special person and the remaining (4) people from the other (9). The number is (^{3}C_{1}\times{}^{9}C_{4}=378).

Correct answer: C. (2541)

Total selections are (^{14}C_{6}). Subtract selections with no named student (^{11}C_{6}) to get (2541).

Correct answer: A. (6952)

The number of senior members can be (2), (3), or (4). The sum is (^{4}C_{2}{}^{12}C_{5}+^{4}C_{3}{}^{12}C_{4}+^{4}C_{4}{}^{12}C_{3}=6952).

Correct answer: B. (272)

Any three of the thirteen points would normally determine a triangle, giving \\(\\binom{13}{3}\\) selections. However, three points on the same straight line do not form a triangle because they are collinear. The five points on the first line create \\(\\binom{5}{3}\\) invalid selections, and the four points on the second line create \\(\\binom{4}{3}\\) invalid selections. The two groups are disjoint, so these invalid cases do not overlap.

Calculate the total: \\(\\binom{13}{3}=286\\). The invalid selections are \\(\\binom{5}{3}=10\\) and \\(\\binom{4}{3}=4\\), for a total of 14. Hence the number of triangles is \\(286-10-4=272\\). The condition about no other three collinear ensures that no further exclusions are needed. Therefore option B is correct.