Combinations

At most (2) means selecting (0), (1), or (2). Hence (\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16).

When exactly 4 sweets are chosen from 9 distinct sweets, order does not matter. Selecting sweets A, B, C, D is the same selection as D, C, B, A. Therefore this is a combination problem, represented by \(\binom{9}{4}\), rather than a permutation. The word “exactly” means that the selection must contain four sweets, not any number up to four.

Using the formula, \(\binom{9}{4}=\frac{9!}{4!5!}\). Cancelling gives \(\frac{9\times8\times7\times6}{4\times3\times2\times1}=126\). Thus the number of selections is 126, so option D is correct. Multiplying in different orders would overcount every group, which is why permutations are not used here.

One employee is already selected, so choose the remaining (4) from (11). The ways are (\binom{11}{4}=330).

Since the fixed member must not be included, only 10 members are available for selection. As the positions in a committee do not matter, use combinations: \(\binom{10}{4}=210\). Hence, the correct answer is 210. Option 330 is incorrect because it equals \(\binom{11}{4}\), which does not exclude the fixed member. Exam tip: When a person must be excluded, first remove that person from the total and then apply combinations.

The two special students are already selected. The remaining (2) students are chosen from (7) in (\binom{7}{2}=21) ways.

Total ways are (\binom{10}{5}=252), and ways with both special players are (\binom{8}{3}=56). Thus the valid ways are (196).

The cases are (2), (3), or (4) girls. The correct count is (\binom{4}{2}\binom{6}{2}+\binom{4}{3}\binom{6}{1}+\binom{4}{4}=115).

This is a combinations problem because a group of three people is selected and the order of its members is irrelevant. There are 10 people in total, so unrestricted selections number C(10,3) = 10!/(3!7!) = 120. The phrase “at least 1 woman” rules out only the selections containing three men. Since there are 5 men, the number of excluded groups is C(5,3) = 10. Thus valid selections are C(10,3) − C(5,3) = 120 − 10 = 110. Therefore option C is correct. Equivalently, the cases of one, two, and three women give C(5,1)C(5,2) + C(5,2)C(5,1) + C(5,3) = 50 + 50 + 10 = 110. Option D is the unrestricted total and includes invalid all-men groups.

For exactly (1) commerce student, the ways are (\binom{3}{1}\binom{7}{2}=3\cdot21=63). Use that exact number when exactly is given.

Using the combination formula, \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\). Thus, \(\binom{18}{1}=\frac{18!}{1!\,17!}=18\). Therefore, option B is correct. The values 36 and 324 are respectively twice and the square of 18, not the value of this combination. Exam tip: remember directly that \(\binom{n}{1}=n\).

Using the combination formula, \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\). Substituting \(n=17\) and \(r=0\), we get \(\binom{17}{0}=\frac{17!}{0!\,17!}=1\), since \(0!=1\). This means there is exactly one way to choose no objects from 17 objects. Option 0 represents the number of objects selected, not the number of ways. Exam tip: Remember that \(\binom{n}{0}=1\) and \(\binom{n}{n}=1\).

In combinations, \(\binom{n}{n}\) means choosing all \(n\) objects from \(n\) objects. There is exactly one way to choose all 11 objects, so \(\binom{11}{11}=1\). The number 11 is the number of objects, not the number of possible selections. Exam tip: remember that \(\binom{n}{n}=\binom{n}{0}=1\).

The combination formula is \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\). Therefore, \(\binom{15}{2}=\frac{15\times14}{2\times1}=105\), so 105 is correct. The value 210 results from not dividing by 2; changing the order of two selected objects does not create a new combination. Exam tip: use \(\binom{n}{2}=\frac{n(n-1)}{2}\) directly.

Combinations follow the symmetry property \(\binom{n}{r}=\binom{n}{n-r}\). Therefore, \(\binom{8}{5}=\binom{8}{8-5}=\binom{8}{3}\), so option B is correct. In \(\binom{8}{2}\), the complementary selection is 6, not 5. Exam tip: Rewrite \(\binom{n}{r}\) as \(\binom{n}{n-r}\) to match such options quickly.

(\binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560). Direct formula is useful for small (r).

Using the complementary property, \\(\binom{n}{r}=\binom{n}{n-r}\\). Hence, \\(\binom{12}{10}=\binom{12}{2}\\). Now, \\(\binom{12}{2}=\frac{12\times11}{2}=66\\), so the sum is \\(66+66=132\\). Option 66 is the value of only one term, so it is a close but incorrect distractor. Exam tip: \\(\binom{n}{r}\\) and \\(\binom{n}{n-r}\\) are always equal.

By Pascal's identity, (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the correct expression is (\binom{7}{3}).

Using the combination formula, \(\binom{n}{2}=\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=45\Rightarrow n(n-1)=90\). Since \(10\times9=90\), \(n=10\). For comparison, \(\binom{9}{2}=36\), not 45. Exam tip: For \(\binom{n}{2}\), first write it as \(\frac{n(n-1)}{2}\).

When order does not matter, the selection is a combination, so choosing 3 students from \(n\) students is \(\binom{n}{3}\). \({}^nP_3\) counts ordered arrangements. Exam tip: use combinations for selection without order.

For the boundary values of combinations, \(\binom{n}{0}=1\) and \(\binom{n}{n}=1\). Therefore, their sum is \(1+1=2\). This is a boundary identity because the number of selected objects is either 0 or all \(n\) objects. Pascal's identity instead has the form \(\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\). Exam tip: Always remember that \(\binom{n}{0}\) and \(\binom{n}{n}\) are both 1.

The direct answer is option B, 10. The special sweet must be included, so we do not choose it again. It is already fixed as one member of the three-sweet selection. We need two more sweets, and there are 5 ordinary sweets available. Therefore the number of choices is C2 = 5!/(2!3!) = (5×4)/2 = 10. Option A, 6, does not count all possible pairs of ordinary sweets. Option B, 10, is correct because each pair added to the special sweet gives one valid selection. Option C, 15, is the total number of ways to choose any 2 sweets from 6, so it ignores the condition that the special sweet must be included. Option D, 20, overcounts and does not follow the required fixed-special-item method. The important idea is that a compulsory item is selected first, leaving fewer items and fewer places to fill. Memory cue: “must include” means fix that item, then choose the remaining items.

Since one toy is broken, it is not available for selection. Thus, 3 toys must be chosen from the remaining 7 good toys. As the order of selection does not matter, the number of ways is \,\(\binom{7}{3}=35\). Option 56 equals \(\binom{8}{3}\), which incorrectly includes the broken toy. Exam tip: use combinations when objects are selected and their order is irrelevant.

There are \(\binom{6}{2}=15\) ways to choose exactly 2 green pencils. The remaining pencil must be blue, which can be chosen in \(\binom{4}{1}=4\) ways. Hence, the total number of selections is \(\binom{6}{2}\binom{4}{1}=15\times4=60\). A value such as 45 results from an incorrect count that does not properly include the required blue pencil. Exam tip: multiply the numbers of ways when choices from separate groups are made independently.

The governing idea is the multiplication principle applied to independent choices from separate colour groups. One red ball can be selected in C(3,1) = 3 ways, one yellow ball in C(4,1) = 4 ways, and one blue ball in C(5,1) = 5 ways. Every red choice can be paired with every yellow choice and every blue choice, so the total is 3 × 4 × 5 = 60. In combination notation this is C(3,1)C(4,1)C(5,1) = 60. Therefore option D is correct. The tempting expression C(12,3) is inappropriate because it permits arbitrary colour distributions, such as three red balls or two red and one yellow. The required one-of-each-colour condition must be enforced by separate choices and multiplication.

The first (2) questions are fixed, so choose the remaining (2) from (10). The ways are (\binom{10}{2}=45).