Combinations

Correct answer: B. (5)

The two special students are already selected. The third student is chosen from the remaining (5) in (\binom{5}{1}=5) ways.

Correct answer: A. (55)

Total ways are (\binom{8}{4}=70), and ways with both special students are (\binom{6}{2}=15). Thus (70-15=55) ways remain.

Correct answer: C. 14

Using \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\), we get \(\binom{14}{1}=\frac{14!}{1!\,13!}=14\). Hence, 14 is correct. Note that \(\binom{14}{0}=1\), so 1 is a common but incorrect distractor here. Exam tip: the number of ways to choose one item is always equal to the total number of available items.

Correct answer: B. 1

Using \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\), we get \(\binom{15}{0}=\frac{15!}{0!\,15!}=1\), since \(0!=1\). There is exactly one way to choose zero objects: choose none. A common confusion is to write 0, but 0 is the number of objects chosen, not the number of possible selections. Exam tip: remember that \(\binom{n}{0}=1\).

Correct answer: B. 1

\(\binom{10}{10}\) means selecting all 10 objects from 10 objects. There is only one way to select every object, so \(\binom{10}{10}=1\). The number 10 represents the objects, not the number of combinations. Exam tip: always remember that \(\binom{n}{n}=1\) and \(\binom{n}{0}=1\).

Correct answer: C. 78

Using \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\), we get \(\binom{13}{2}=\frac{13!}{2!11!}=\frac{13\times12}{2}=78\). Hence, option C is correct. \(156\) results if \(13\times12\) is not divided by \(2!\). Exam tip: use \(\binom{n}{2}=\frac{n(n-1)}{2}\) directly for selections of two objects.

Correct answer: C. 495

\(\binom{12}{4}=\frac{12!}{4!\,8!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\). Therefore, 495 is the correct answer. The value 11880 is \(^{12}P_4\), where different orders are counted separately. Exam tip: In combinations, order does not matter, so divide by \(r!\).

Correct answer: A. \(\binom{9}{4}\)

Combinations satisfy the symmetry property \(\binom{n}{r}=\binom{n}{n-r}\). Therefore, \(\binom{9}{5}=\binom{9}{9-5}=\binom{9}{4}\). \(\binom{9}{3}\) has a different number of selected objects, so it is not equal to the given expression. Exam tip: You can quickly replace \(\binom{n}{r}\) with \(\binom{n}{n-r}\).

Correct answer: C. 42

\(\binom{7}{2}=\frac{7\times6}{2\times1}=21\). Using the complementary property \(\binom{n}{r}=\binom{n}{n-r}\), \(\binom{7}{5}=\binom{7}{2}=21\). Hence, \(21+21=42\), so option C is correct. The value \(35\) is \(\binom{7}{3}\), not \(\binom{7}{5}\). Exam tip: in \(\binom{n}{r}\), use the smaller of \(r\) and \(n-r\) to calculate quickly.

Correct answer: B. 7

We use \(\binom{n}{2}=\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=21\Rightarrow n(n-1)=42\). Since \(7\times6=42\), \(n=7\). For example, \(\binom{6}{2}=15\), so 6 is not correct. Exam tip: For questions involving \(\binom{n}{2}\), first write it as \(\frac{n(n-1)}{2}\).

Correct answer: B. 15

Using the combination identities, \(\binom{n}{1}=n\) and \(\binom{n}{0}=1\). Thus, the given equation becomes \(n+1=16\), which gives \(n=15\). If \(n=16\), the sum would be 17, so it is not correct. Exam tip: Remember directly that \(\binom{n}{0}=1\) and \(\binom{n}{1}=n\).

Correct answer: C. 30

\(\binom{6}{2}=\frac{6!}{2!4!}=15\). Also, \(\binom{6}{4}=\binom{6}{2}=15\), using \(\binom{n}{r}=\binom{n}{n-r}\). Therefore, \(\binom{6}{2}+\binom{6}{4}=15+15=30\). Option 15 is the value of only one term, not their sum. Exam tip: use \(\binom{n}{r}=\binom{n}{n-r}\) for complementary selections.

Correct answer: A. (\binom{6}{3})

Direct answer: Option A, \(\binom{6}{3}\). Pascal's identity says \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Compare the expression with this pattern: take n=5 and r=2. Then \(\binom{5}{2}+\binom{5}{3}=\binom{6}{3}\). Numerically, the left side is 10+10=20, and \(\binom{6}{3}=20\), which confirms the identity. Option A is correct. Option B, \(\binom{6}{2}\), equals 15, so it does not match. Option C, \(\binom{5}{5}\), equals 1 and is unrelated to the required adjacent pair. Option D, \(\binom{10}{5}\), equals 252 and comes from a different combination problem. The important feature is that the lower numbers 2 and 3 are consecutive, and the upper number increases from 5 to 6. Memory cue: adjacent combinations in one row combine into the next row of Pascal's triangle.

Correct answer: C. 63

For each of the 6 distinct coins, there are two choices: select it or do not select it. Hence, the total number of selections, including the empty selection, is \(2^6=64\). Since at least 1 coin must be selected, exclude the empty selection: \(64-1=63\). Option 64 is incorrect because it also includes selecting no coin. Exam tip: for “at least one” selection questions, subtract the empty selection from the total number of subsets.

Correct answer: D. 32

For each of the 5 distinct cards, there are two independent choices: select it or do not select it. Therefore, the total number of selections is \(2^5=32\). This includes the one way of selecting no card, because the question says “any number.” The answer would be \(31\) only if selecting at least one card were required. Exam tip: “any number” usually includes 0; subtract the empty selection only when the question says “one or more.”

Correct answer: C. 11

The phrase “at least 2” includes selections of exactly 2, exactly 3, or exactly 4 fruits. Since the order of selected fruits does not matter, use combinations: C(4,2)+C(4,3)+C(4,4)=6+4+1=11. Therefore option C is correct. A complementary method confirms the result. Four fruits have 2^4=16 total subsets, including the empty selection and the four selections containing exactly one fruit. These five cases are not allowed, so the required number is 16−1−4=11. Option A counts only pairs and omits selections of three and four fruits. Option B omits one category, while option D counts all subsets, including the forbidden selections of zero or one fruit.

Correct answer: B. 8

Selecting at most 1 subject includes two cases: selecting no subject and selecting exactly 1 subject. Thus, the total number of ways is \(\binom{7}{0}+\binom{7}{1}=1+7=8\). Option 7 counts only the selections of exactly one subject and misses the one way of selecting none. Exam tip: “At most” includes every possible selection below the stated number as well.

Correct answer: B. 36

Since the order of selection does not matter, use combinations. Thus, the number of ways is \(\binom{9}{2}=\frac{9\times8}{2\times1}=36\). Choosing 72 would count different orders separately, which is not required here. Exam tip: use \(nCr\) whenever order does not affect a selection.

Correct answer: C. (210)

Letters are only being selected, so (\binom{10}{4}=210). Use combination when arrangement is not asked.

Correct answer: B. 15

The governing concept is combination because the question asks for selections, not an arrangement of the chosen pictures in named positions. Choose 2 pictures from 6 using 6C2 = 6!/(2!4!) = (6×5)/(2×1) = 15. Therefore option B is correct. A selection containing pictures P and Q is the same as one written Q and P, so reversing their order must not create a new selection. Option A does not follow the combination formula. Option C is the un-divided product 6×5 and therefore counts both orders of each pair. Option D, 6², represents ordered choices with repetition in a different model. If the poster had two distinct positions, an arrangement count would be needed, but no such positions are specified.

Correct answer: C. \(35\)

Selecting \(3\) flavors from \(7\) flavors without order is a combination:
\[\binom{7}{3}=\frac{7!}{3!\,4!}=\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1}=35.\]
So the number of ways is \(35\).

Correct answer: B. 56

This is a selection problem because the order in which the tickets are chosen does not matter. Hence, the number of ways is \(\binom{8}{5}=\frac{8!}{5!3!}=56\). We can also use \(\binom{8}{3}\), since choosing 5 tickets is equivalent to leaving out 3 tickets. Exam tip: use combinations \(\binom{n}{r}\) when order is not important.

Correct answer: A. 3

The governing concept is conditional combination: the requirement that both selected books must be Hindi excludes every English book from the choice. Thus we only choose 2 books from the 3 Hindi books, giving 3C2 = 3!/(2!1!) = (3×2)/2 = 3. Therefore option A is correct. The four English books are irrelevant because no English book can appear in a valid selection. Option B is the product 3×2 before removing the duplicate order of each pair. Option C incorrectly combines the two categories, and option D does not represent the stated restriction. Since a pair of Hindi books has no order, permutations are not used.

Correct answer: C. 30

This is a combination problem because only the members selected matter; their order within the senior group or junior group does not matter. Choose 2 of the 4 senior students in C(4,2) = 4!/(2!2!) = 6 ways. Independently, choose 1 of the 5 junior students in C(5,1) = 5 ways. By the multiplication principle, every senior selection can be paired with every junior selection, so the total is C(4,2) × C(5,1) = 6 × 5 = 30. Therefore option C is correct. Options A, B, and D result from incorrectly adding groups or using an inappropriate ordered-selection calculation.

Correct answer: C. 100

Because only a group is being selected and not arranged, combinations are required. Choose one science student from five in C(5,1) = 5 ways. Next choose three arts students from six in C(6,3) = 6!/(3!3!) = 20 ways. These two choices are independent: each science selection can be paired with every valid arts selection. By the multiplication principle, the total is C(5,1) × C(6,3) = 5 × 20 = 100. Therefore option C is correct. Adding the two counts would not form a complete four-student group, and no permutation factor is needed because rearranging the selected students does not create a new selection. Options A, B, and D do not result from the required combination product.