Combinations

Only a selection of 2 sweets from 6 is required, so the order of selection does not matter. Hence, the number of ways is \(\binom{6}{2}=\frac{6\times5}{2\times1}=15\). The value 30 would result if order were counted, using the permutation \(^{6}P_2\). Exam tip: When a question says “choose” or “select” and does not mention order, use combinations.

Since the positions of the colours are not fixed, choosing red and blue is the same as choosing blue and red. Thus, order does not matter and combinations are used: \(\binom{9}{2}=\frac{9\times8}{2}=36\). Therefore, the correct answer is 36. The value 72 comes from the permutation \(9\times8\), where different orders are counted separately. Exam tip: use \(\binom{n}{r}\) when only selection matters and order is irrelevant.

There is exactly 1 way to choose 0 gifts: choose none. There are \(\binom{10}{1}=10\) ways to choose 1 gift. Hence, the total is \(\binom{10}{0}+\binom{10}{1}=1+10=11\). Choosing 10 alone is incorrect because it omits the one way of choosing no gift. Exam tip: Add the counts for mutually exclusive selection cases joined by “or.”

Here, 4 chapters are to be selected from 7 chapters, and the order of selection does not matter. Therefore, use combinations: \(\binom{7}{4}=\frac{7!}{4!3!}=35\). Hence, the correct answer is 35. Option 210 is \(7P4\), which applies when order matters, so it is not correct here. Exam tip: use \(nCr\) when objects are selected without considering their order.

We need to select 2 questions out of 11 to be left out. Since the order in which the questions are selected does not matter, use combinations: \(\binom{11}{2}=\frac{11\times10}{2\times1}=55\). Therefore, the correct answer is 55. The value 110 would result if order were counted, that is, if permutations were used. Exam tip: use \(\binom{n}{r}\) when items are selected and their order is not specified.

If 9 students are selected from 12, then 3 students are left unselected. Hence, \(\binom{12}{9}=\binom{12}{12-9}=\binom{12}{3}\). Option 9 represents the number selected, not the number unselected. Exam tip: use the identity \(\binom{n}{r}=\binom{n}{n-r}\) in combination questions.

The governing concept is combinations with a complementary condition. Because a selection is a group, not an arrangement, order does not matter. There are 8 students altogether, so the total number of selections of 3 is C(8,3) = 56. The condition “at least 1 girl” excludes only selections containing no girl, that is, selections of 3 boys from the 5 boys. Their number is C(5,3) = 10. Hence the required number is 56 − 10 = 46. Direct verification gives C(3,1)C(5,2) + C(3,2)C(5,1) + C(3,3)C(5,0) = 30 + 15 + 1 = 46. Therefore option C is correct. Option D counts all selections and wrongly includes the ten all-boy groups; the other values do not satisfy the condition.

This is a selection problem, so the order of the badges does not matter. Hence, the number of ways is \(\binom{18}{2}=\frac{18\times17}{2}=153\). The value \(306\) would result if the order of the two badges were also counted. Exam tip: When a question says “choose” or “select” and no order is specified, use combinations.

Selecting a duty team means choosing five volunteers, with no positions or order specified. Therefore this is a combination problem. The number of ways is \(\binom{10}{5}=\frac{10!}{5!5!}\). Cancelling the factorials or using the product form gives \(\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252\).

Order is not counted separately: the same five volunteers form one team regardless of which name is written first. Using \(10!\) would count arrangements rather than teams and would greatly overcount. Thus the number of possible duty teams is 252, which is option B. The equality \(\binom{10}{5}=\binom{10}{5}\) also reflects that choosing five members is equivalent to leaving the other five out.

Here, only 3 chapters are to be chosen from 13 different chapters, so the order of selection does not matter. Therefore, use combinations: \(\binom{13}{3}=\frac{13\times12\times11}{3\times2\times1}=286\). Hence, option C is correct. Option D, \(13\times12\times11=1716\), counts different orders as well, so it represents a permutation. Exam tip: Use \(nCr\) when only selection is involved and order is not specified.

This is a selection problem, not an arrangement problem, so combinations are used. The required number of ways is \(\binom{9}{6}=\frac{9!}{6!3!}=84\). Equivalently, \(\binom{9}{6}=\binom{9}{3}\), because choosing 6 files is the same as leaving out 3 files. The nearby distractor \(126\) is the value of \(\binom{9}{4}\), not \(\binom{9}{6}\). Exam tip: use \(nCr\) when the order of selection does not matter.

The boys and girls form two separate groups, and the required selection must contain exactly three boys and exactly one girl. First choose the boys, then choose the girl. These choices are independent, so their numbers are multiplied. Since selection is involved and order is irrelevant, combinations are used.

The number of ways to choose three boys is \(\binom63=20\). The number of ways to choose one girl is \(\binom51=5\). Therefore the total is \(\binom63\binom51=20\times5=100\). The order in which the selected students are listed does not create a new group. Hence choice A, (100), is correct.

The governing concept is combination because only selection matters; the order in which flowers are chosen does not matter. First select 2 red flowers from 8: C(8,2) = 8×7/(2×1) = 28. Independently, select 2 white flowers from 7: C(7,2) = 7×6/(2×1) = 21. Since both conditions must occur together, apply the multiplication principle: C(8,2)×C(7,2) = 28×21 = 588. Therefore, option B is correct. Option A does not use the full product correctly, option C overcounts the selections, and option D is much too small. Flowers of the same colour are not ordered, so permutations are unnecessary.

Two doctors can be selected in \(\binom{7}{2}=21\) ways, and one nurse can be selected in \(\binom{4}{1}=4\) ways. Since these selections are independent, the multiplication principle gives \(\binom{7}{2}\binom{4}{1}=21\times4=84\) teams. The option \(56\) does not correctly combine the selections of doctors and nurses. Exam tip: when fixed numbers are chosen from separate groups, multiply the corresponding combinations.

The two kinds of books must be selected separately because the requirement is exactly two Hindi books and exactly three Sanskrit books. For the Hindi books, choose 2 from 5, giving \\(\\binom{5}{2}=10\\) possibilities. For the Sanskrit books, choose 3 from 6, giving \\(\\binom{6}{3}=20\\) possibilities. The order in which books are selected does not matter, so combinations are appropriate rather than permutations.

A Hindi selection can be paired with any Sanskrit selection, so the multiplication principle gives \\(\\binom{5}{2}\\binom{6}{3}=10\times20=200\\). Therefore option D is correct. Adding the two counts would be inappropriate because every valid result includes both a Hindi group and a Sanskrit group. The product counts all possible pairs of the two independent selections exactly once.

Because the posts are distinct, order matters. There are 8 choices for chairperson and then 7 remaining choices for secretary, so \(8\times7=56\). \(\binom{8}{2}=28\) counts an unordered pair only. Exam tip: use permutations when positions are different.

Each match is determined by a pair of distinct teams. Therefore, the number of matches equals the number of ways to choose 2 teams from 11: \(\binom{11}{2}=\frac{11\times10}{2}=55\). Hence, option B is correct. Counting 110 would count every match twice, such as A versus B and B versus A. Exam tip: for a single round-robin league with \(n\) teams, total matches are \(\binom{n}{2}\).

A triangle is formed by selecting any 3 of the 14 points. Since no 3 points are collinear, every selected group of 3 points forms a triangle. Therefore, the number of triangles is \(\binom{14}{3}=\frac{14\times13\times12}{3\times2\times1}=364\). The value 1092 can result from an arrangement-based calculation, but the order of vertices does not matter for a triangle. Exam tip: use combinations when objects are selected and their order is irrelevant.

A straight line is determined by a pair of distinct points. Since no three points are collinear, every pair of points gives a different line. Therefore, the number of lines is \(\binom{12}{2}=\frac{12\times11}{2}=66\). The value \(132\) would count each pair in two orders. Exam tip: when no three of \(n\) points are collinear, use \(\binom{n}{2}\) to find the number of lines.

Any three non-collinear points determine one unique circle. Since no three of the given points are collinear, any set of 3 points can be chosen. Therefore, the number of circles is \(\binom{10}{3}=\frac{10\times9\times8}{3\times2\times1}=120\). The value 90 does not represent the required unordered selection; the order of chosen points is irrelevant here. Exam tip: Use combinations \(\binom{n}{r}\) when the order of selection does not matter.

Only a selection of 4 guests from 8 is required; the order of calling them is not important. Hence, use combinations: \(\binom{8}{4}=\frac{8!}{4!\,4!}=70\). The value 1680 is \(8P_4\), which would apply if the four guests were called in a specific order. Exam tip: use \(nCr\) when order does not matter.

This is a selection question, so combinations are used: \(\binom{16}{4}=\frac{16!}{4!\,12!}=1820\). Therefore, there are 1820 ways to choose 4 cards from 16 cards. \(43680\) is the value of the permutation \(^{16}P_4\), where the order of the cards matters; no order is asked for here. Exam tip: When a question says “choose” or “select” without mentioning order, usually use \(nCr\).

When 3 books are selected, changing their order gives the same group, so it is a combination. In ranks, offices and passwords, changing the order changes the outcome. Exam tip: use combinations when only selection matters, not arrangement.

Each coin has two choices: it may be selected or not selected. Hence, the total number of subsets of 8 distinct coins is \(2^8=256\). One of these is the empty selection, in which no coin is chosen. Therefore, the required number is \(256-1=255\). Thus, 255 is correct; 256 incorrectly includes the empty selection. Exam tip: for “at least one” selection questions, subtract the empty selection from the total.

Each of the 6 colours has two possibilities: selected or not selected. Hence, there are \(2^6=64\) total subsets. The 1 selection with no colour chosen and the \(\binom{6}{1}=6\) selections with exactly one colour chosen are not allowed. Therefore, the required number is \(64-1-6=57\). Option 63 excludes only the empty selection, but selections of exactly one colour must also be excluded. Exam tip: For “at least” questions, counting the complement is often quicker.