Combinations

There are 6 ways to choose a pencil, and for each choice there are 2 ways to choose a notebook. By the multiplication principle, the total number of ways is \(6\times2=12\). The values 8 and 10 result from addition, but both choices are made together, so their possibilities must be multiplied. Exam tip: multiply the numbers of ways for independent successive choices.

By the multiplication principle, the subject can be chosen in 3 ways and the game in 5 ways, so the total number of selections is \(3 \times 5 = 15\). The answer 8 would result from adding the choices, but both selections are made together, so multiplication is required. In an exam, use the multiplication rule when independent choices are made successively.

The traveller can choose one of the 4 bus routes or one of the 3 train routes. Since these choices are mutually exclusive, the addition principle gives \(4+3=7\) ways. Multiplying to get 12 would be incorrect because both a bus route and a train route are not being selected together. Exam tip: add mutually exclusive ‘or’ choices and multiply successive ‘and’ choices.

The student can choose one of the 6 short-answer questions or one of the 4 long-answer questions. These are mutually exclusive alternatives, so the addition rule applies: 6 + 4 = 10. The value 24 (= 6 × 4) would apply if the student had to choose both one short and one long question. Exam tip: add the counts for ‘or’ choices and multiply for combined ‘and’ choices.

By the multiplication principle, the numbers of independent choices are multiplied: \(2\times3\times2=12\). Therefore, the correct answer is 12. The distractor 6 results from multiplying only 2 and 3 and ignoring the 2 dessert choices. Exam tip: When one choice is made from each of several independent categories, multiply the number of choices in all categories.

There are 3 choices for the pen, 4 choices for the marker, and 2 choices for the pencil. Since these choices are independent and all three items must be selected, the multiplication principle gives \(3\times4\times2=24\) ways. The value 12 fails to account correctly for all three selection stages. Exam tip: multiply the numbers of choices when independent choices are connected by ‘and’.

By the multiplication principle, the choices for all four positions are multiplied. Thus, the total number of vehicle numbers is \\(4\times4\times10\times10=1600\\). Since repetition is allowed, each letter and digit position retains its full set of choices. The answer 400 does not correctly multiply the choices for all four positions. Exam tip: multiply the number of choices available at each successive independent position.

This is a combination question because the order in which the books are chosen does not matter. Hence, the number of selections is \(\binom{5}{2}=\frac{5\times4}{2\times1}=10\). Option 20 is obtained using permutation \(5P2\), where order is counted. Exam tip: use combinations for selection problems when order is not important.

The order of students does not matter when forming a team, so we use combinations. The required number is \(\binom{6}{3}=\frac{6!}{3!\,3!}=20\). Option 18 is incorrect because it is not the number of combinations of choosing 3 students from 6. Exam tip: Use \(\binom{n}{r}\) for selection, teams, or groups when order does not matter.

This is a selection of 2 flowers from 7 different flowers where order does not matter. Therefore, the number of ways is \(\binom{7}{2}=\frac{7!}{2!\,5!}=\frac{7\times6}{2}=21\). The value \(42\) would arise if order mattered, making it an arrangement question. Exam tip: For words such as “choose” or “select,” usually use combinations, \(nCr\).

A group has no order, so (\binom{8}{4}=70). In exams, the word group usually indicates combination.

When exactly one question is selected from 9 questions, each question gives one distinct selection. Therefore, the number of ways is \(\binom{9}{1}=9\). Option 8 is incorrect because it represents the number of ways to leave out one question, not to select one question. Exam tip: apply the identity \(\binom{n}{1}=n\) directly.

When 0 pens are selected, the selected group is the empty set. Leaving out all 10 pens always gives this same single group; hence \(\binom{10}{0}=1\). Option B is incorrect because excluding different individual pens does not create different selections when no pen is selected. Exam tip: In general, \(\binom{n}{0}=\binom{n}{n}=1\).

In a combination, the order of chosen objects is ignored; choosing A and B is the same as choosing B and A. If order matters, it is a permutation. Exam tip: the phrase “order matters” indicates permutation.

The order of the two friends does not matter when forming a pair, so we use combinations: \(\binom{6}{2}=\frac{6\times5}{2\times1}=15\). Hence, the correct answer is 15. The value 30, \(6\times5\), counts order separately and therefore counts each pair twice. Exam tip: use \(\binom{n}{r}\) whenever selection is made without regard to order.

This is a combination problem because selection is involved and the order in which the chosen boy and girl are named does not create a new selection. Choose one boy from four in C(4,1) = 4 ways, and choose one girl from three in C(3,1) = 3 ways. Since these choices come from independent groups, apply the multiplication principle: C(4,1) × C(3,1) = 4 × 3 = 12. Therefore option B is correct. The value 7 merely adds the group sizes and does not count pairs; 24 counts an inappropriate ordering; and 1 ignores the different possible choices.

Only the five boys are involved because the required group must contain two boys and no girls. Since a selection is a group and not an ordered arrangement, choosing Ravi and Aman is the same outcome as choosing Aman and Ravi. Therefore combinations are appropriate. The number of ways is C(5,2) = 5!/[2!3!] = (5 × 4)/(2 × 1) = 10. Thus option A is correct. If order had mattered, the calculation would have been P(5,2) = 5 × 4 = 20, which explains option B as a permutation-based distractor. The total of nine students is irrelevant because the question restricts the selection to boys only; options C and D have no valid counting basis.

The apple selection and mango selection are independent, so count them separately and multiply. One apple can be selected from 4 apples in \(\binom41=4\) ways. Two mangoes can be selected from 5 mangoes in \(\binom52=\frac{5\times4}{2\times1}=10\) ways. By the multiplication principle, the total is \(4\times10=40\).

This is a combination question because only the selected fruits matter; the order in which they are chosen does not matter. For example, choosing mango X and then mango Y is the same selection as choosing Y and then X, so permutations must not be used. Therefore the number of possible selections is 40, which is option C.

A line is determined by choosing 2 points. Therefore, the number of lines is \(\binom{8}{2}=\frac{8\times7}{2}=28\). Order does not matter: choosing A and B is the same as choosing B and A. The value 56 comes from \(8\times7\), which counts every pair twice. Exam tip: use combinations \(\binom{n}{r}\) when order is not important.

A triangle needs (3) points, so (\binom{7}{3}=35). The non-collinear condition makes triangles possible.

A circle is uniquely determined by three non-collinear points. Since no three of the given points are collinear, every selection of 3 points from 6 forms one circle. Therefore, the number of circles is \(\binom{6}{3}=20\). The value \(\binom{6}{2}=15\) counts pairs of points or line segments, not circles. Exam tip: For counting circles from points, choose sets of three points.

A handshake forms a pair of two persons, and the order of the two people does not matter. Hence, the total number of handshakes is \(\binom{10}{2}=\frac{10\times9}{2}=45\). The value 90 can result from counting ordered pairs, which counts each handshake twice. Exam tip: For handshake questions, usually use \(\binom{n}{2}\).

The direct answer is option C: 84 committees. A committee is a group, not an ordered arrangement. Therefore choosing A, B, C is the same committee as choosing C, B, A. We select 3 members from 9 without assigning posts, so use combinations: \(\binom{9}{3}=\frac{9\times8\times7}{3\times2\times1}=84\). Option C is correct. Option A, 27, is not the value of this combination. Option B, 72, also does not result from the correct formula. Option D, 504, is \({}^{9}P_3\), which counts orders and therefore counts each committee six times. Since \(3!=6\), \(504\div6=84\). The absence of named positions is the key fact. Memory cue: committee or team selection without roles means combination; arrangement or post assignment means permutation.

Only a group of 2 candidates is to be selected, so the order of selection does not matter. Therefore, use combinations: \(\binom{12}{2}=\frac{12\times11}{2\times1}=66\). Hence, the correct answer is 66. The value 132 would result from counting different orders of the same two candidates separately, which is not required here. Exam tip: Questions about selecting a team, committee, or group generally use combinations, \(\binom{n}{r}\).

Here, 3 cards are selected from 11 distinct cards, and the order of selection does not matter. Therefore, the number of combinations is \(\binom{11}{3}=\frac{11\times10\times9}{3\times2\times1}=165\). The value 990 is related to permutations, where the order of selection is also counted. Exam tip: When a question asks only to “choose” or “select,” it usually requires combinations.