Combinations

Only 3 pictures are being selected, so the order in which they are displayed does not matter. Hence, use combinations: \(\binom{8}{3}=\frac{8!}{3!\,5!}=56\). Therefore, the correct answer is 56. The value 336, \(8\times7\times6\), would arise from permutations when order is considered important. Exam tip: When a question says “choose” or “select” and no order is specified, use combinations.

The governing idea is combination, since the question explicitly says that order is not considered. Choose 2 objects from 6 by using 6C2 = 6!/(2!4!). This becomes (6×5)/(2×1) = 30/2 = 15, so option B is correct. For example, choosing digits a and b is the same selection as choosing b and a; they must not be counted twice. Option A is only 6×2 and does not apply the combination rule. Option C is the un-divided product 6×5, which counts both orders. Option D, 6², represents a different type of ordered choice with repetition allowed, not this selection problem.

A line segment is determined by choosing any 2 of the 10 points. Hence, the number of segments is \(\binom{10}{2}=\frac{10\times9}{2}=45\). The value \(90\) counts ordered pairs, but segments AB and BA are the same. Exam tip: for questions involving the selection of two points, usually use \(\binom{n}{2}\).

Changing the order of the same 5 students does not change a committee, so combinations apply. In ranks, offices and passwords, order creates a different outcome. Exam tip: use combinations for selections without designated roles.

While selecting a committee, the order of the chosen students does not create a new committee, so combinations apply. In option B, positions are different, making order important. Exam tip: words such as “select” and “committee” usually indicate combinations.

This is a selection problem, so the order of the toys does not matter. Hence, the number of ways is \(\binom{12}{4}=\frac{12!}{4!\,8!}=495\). The value \(11880\) is \(^{12}P_4\), which counts arrangements where different orders of the chosen toys are treated separately. Exam tip: Words such as “choose” or “form a committee” usually indicate combinations.

For each of the 7 pencils, there are two choices: select it or do not select it. Thus, the total number of selections, including the empty selection, is \(2^7=128\). Since at least one pencil must be selected, exclude the empty selection: \(128-1=127\). Option 128 incorrectly includes selecting no pencil. Exam tip: For selections of at least one item from \(n\) distinct items, use \(2^n-1\).

While selecting a pair from 15 students, the order of the two students does not matter. Therefore, the number of ways is
\(\binom{15}{2}=\frac{15\times14}{2\times1}=105\). Hence, 105 is correct. The value 210 counts each pair in two orders and is related to permutations. Exam tip: use combinations when only selection or grouping is involved.

Direct answer: Option B, 28. A team is a selection, so order does not matter. Choosing six students from eight is therefore a combination, written as \(\binom{8}{6}\). Using the formula, \(\binom{8}{6}=\frac{8!}{6!2!}=\frac{8×7}{2×1}=28\). An even easier idea is to choose the two students who are left out: \(\binom{8}{2}=28\), and every six-person team corresponds to exactly one excluded pair. Option A, 14, is too small and does not count all selections. Option B, 28, is correct. Option C, 48, does not follow the combination calculation. Option D, 56, is another overcount, often caused by treating selections as ordered or using the wrong factorial expression. Since a team has no first, second, or third member, permutations are not needed. Memory cue: selection without order means combination.

The order of the 3 selected books does not matter, so we use combinations. The required number is \(\binom{9}{3}=\frac{9!}{3!\,6!}=84\). Option 504 is \(^{9}P_{3}\), which counts arrangements where order matters, so it is not correct. Exam tip: words such as selection, group, or set usually indicate combinations.

This is a selection problem, so the governing concept is combination rather than permutation. A practice group has no first, second, third, or fourth position; the same four players form one group regardless of their listing order. Therefore, calculate 11C4 = 11!/(4!7!) = (11×10×9×8)/(4×3×2×1) = 7920/24 = 330. Thus option C is correct. Option A is only 11×4 and ignores the selection structure. Option B is not the value of 11C4. Option D is the numerator before division by 4!, so it counts arrangements of each group and overcounts every group 24 times. The supplied answer is corrected to C.

This is a selection question, so the order of the pens does not matter. Hence, the number of ways is \(\binom{12}{3}=\frac{12\times11\times10}{3\times2\times1}=220\). Therefore, 220 is correct. The value 440 can result from unnecessarily counting the same selections twice. Exam tip: use combinations \(\binom{n}{r}\) when objects are selected and their order is not specified.

This is a selection problem, so the order of the chosen questions does not matter. Therefore, the number of ways is \(\binom{10}{7}=\binom{10}{3}=120\). Option 5040 equals \(7!\) and relates to arranging seven selected questions, not selecting them. Exam tip: Use combinations \(\binom{n}{r}\) when the question asks only to choose items.

This problem uses combinations together with the multiplication principle. First choose the required two boys from five: C(5,2) = 5!/(2!3!) = 10. Independently, choose two girls from four: C(4,2) = 4!/(2!2!) = 6. Every valid pair of boys can be combined with every valid pair of girls, so the total number of four-student groups is the product 10 × 6 = 60. Therefore option C is correct. Addition would be inappropriate because both selections must occur together. Option A omits the girls’ choices, option B does not equal the required product, and option D overcounts by treating the selected students as ordered or by using an incorrect arrangement method.

The two colors are separate groups, so first choose two red balls from six and then choose one blue ball from five. The red selection can be made in \(\binom62=\frac{6\times5}{2}=15\) ways. The blue selection can be made in \(\binom51=5\) ways. Since every red choice can be combined with every blue choice, multiply them: \(15\times5=75\).

The balls of the same color are selected without regard to order, so combinations are appropriate. We do not arrange the selected balls, because the question asks only which balls are chosen. Therefore the total number of selections is 75, which is option D. A result such as 30 would undercount the possible pairs of red balls.

The governing concept is combination with independent category requirements. First select 1 man from 7, which can be done in 7C1 = 7 ways. Then select 2 women from 3, which can be done in 3C2 = 3!/(2!1!) = 3 ways. By the multiplication principle, the total number of complete selections is 7C1 × 3C2 = 7×3 = 21. Hence option B is correct. Option A adds the group sizes instead of combining the two required choices. Option C doubles the correct result and does not follow either combination or multiplication calculation. Option D is 3 times the correct result. Since only membership of the selected group matters, no permutation factor is needed.

The governing concept is combination selection from separate categories, followed by multiplication. One mathematics book can be chosen in 4C1 = 4 ways. Two physics books can be chosen in 5C2 = 5!/(2!3!) = (5×4)/2 = 10 ways. Every mathematics choice can be paired with every valid physics pair, so the multiplication principle gives 4C1 × 5C2 = 4×10 = 40. Thus option C is correct. Option A does not count the physics pairs, option B is an incomplete or incorrect product, and option D results from overcounting or using an inappropriate ordered calculation. The books within each subject are selected as groups, so their internal order is irrelevant.

A line segment has two endpoints. Hence, the number of line segments equals the number of ways of choosing 2 points from 13 points: \(\binom{13}{2}=\frac{13\times12}{2}=78\). Therefore, 78 is correct. The value 156 counts each pair in two orders, but the order of endpoints does not matter for a line segment. Exam tip: use \(\binom{n}{2}\) when choosing two items and order does not matter.

To form a triangle, choose any 3 of the 8 points. Since no 3 points are collinear, every such selection forms a triangle. Therefore, the number of triangles is \(\binom{8}{3}=\frac{8\times7\times6}{3\times2\times1}=56\). The option 112 may result from incorrectly counting arrangements, but the order of selecting points does not matter. Exam tip: Use combinations \(\binom{n}{r}\) when the order of selection is irrelevant.

A unique circle is determined by three non-collinear points. Since no three of the given points are collinear, every selection of 3 points from 9 gives one circle. Thus, the number of circles is \(\binom{9}{3}=\frac{9\times8\times7}{3\times2\times1}=84\). Option 126 equals \(\binom{9}{4}\), so it does not apply here. Exam tip: for circles, choose 3 non-collinear points.

Each handshake corresponds to one pair of distinct persons, and each pair is counted only once. Therefore, the total number of handshakes is \(\binom{14}{2}=\frac{14\times13}{2}=91\). The value 182 results from counting every pair twice. Exam tip: for handshake questions, usually use \(\binom{n}{2}\).

Each match is played between a pair of distinct teams. Therefore, the number of matches is \,\(\binom{16}{2}=\frac{16\times15}{2}=120\). Getting 240 would count every pair twice. Exam tip: For “each team plays every other team once” questions, usually use \(\binom{n}{2}\).

Since no positions such as president or secretary are assigned, the order of selected members does not matter. Therefore, the number of ways is \(\binom{6}{3}=\frac{6!}{3!3!}=20\). The value 120 is \(^{6}P_3\), which counts arrangements where order matters, so it is not applicable here. Exam tip: use combinations when only selection is involved and no posts are assigned.

The fixed member is already selected, so choose the remaining (3) from (9). The ways are (\binom{9}{3}=84).

Since the fixed member cannot be included, only 8 members are available for selection. The order of members in a committee does not matter, so use combinations: \(\binom{8}{3}=\frac{8\times7\times6}{3\times2\times1}=56\). Hence, the correct answer is 56. Option 84 equals \(\binom{9}{3}\), which incorrectly includes the excluded member among the available people. Exam tip: use \(nCr\) when forming a committee because order is irrelevant.