Combinations

The governing principle is the product rule together with combinations, because the required selection is split into two disjoint groups. The multiples of 6 from 1 through 48 are 6, 12, ..., 48, so there are 48/6 = 8 such numbers. The remaining 48 - 8 = 40 numbers are non-multiples of 6. We must choose 4 from the first group and 2 from the second, without regard to order. Therefore the number of selections is C(8,4) × C(40,2). Now C(8,4) = 70 and C(40,2) = 40×39/2 = 780, so the product is 70 × 780 = 54,600. Hence option D is correct. Permutations would overcount each selection because rearranging its members does not create a new selection.

The selection is unordered, so combinations are required. Between 1 and 35 inclusive, there are 18 odd numbers and 17 even numbers: the odd numbers are 1, 3, ..., 35, while the evens are 2, 4, ..., 34. Choose four odds in C(18,4) ways and two evens in C(17,2) ways. By the multiplication principle, the total is C(18,4)C(17,2) = 3,060 × 136 = 416,160. Hence option C is correct. Adding the two combination counts would be inappropriate because both choices must be made together; permutations would overcount different orders of the same selected set.

After removing (7) special students (15) students remain. Hence there are (\binom{15}{8}=6435) ways.

The group can contain 4, 5, 6, or 7 teachers. Therefore, the total number of ways is \(\binom{11}{4}\binom{13}{3}+\binom{11}{5}\binom{13}{2}+\binom{11}{6}\binom{13}{1}+\binom{11}{7}\binom{13}{0}\) = \(94380+36036+6006+330=136752\). Hence, 136752 is correct. Option 94380 counts only the case of exactly 4 teachers, not every case covered by “at least 4”. Exam tip: For “at least,” add all cases from the stated number up to the maximum possible number.

Total ways are (\binom{21}{7}=116280). Removing all-white (\binom{10}{7}=120) and all-black (\binom{11}{7}=330) gives (115830).

Since all prizes are identical, the order of the 6 selected candidates does not matter. Hence, use combinations: \(\binom{16}{6}=\frac{16!}{6!\,10!}=8008\). The value 5765760 corresponds to the permutation \(^{16}P_6\), which would count arrangements if the prizes were distinct. Exam tip: use \(nCr\) when only selection matters and order is irrelevant.

Total ways are (\binom{13}{6}=1716) and if no special friend goes then (\binom{8}{6}=28). Hence (1716-28=1688).

Choose (3) from the special fruits and (3) from the remaining (8). The ways are (\binom{6}{3}\binom{8}{3}=1120).

The governing idea is independent combination followed by multiplication. The red balls can be chosen in C(9,2) ways, the blue balls in C(10,2) ways, and the green balls in C(11,2) ways. Since a complete selection requires all three color choices, multiply them: C(9,2)C(10,2)C(11,2) = 36 × 45 × 55 = 89,100. Therefore option D is correct. Adding the three counts would describe a different situation, and using permutations would count ordering that is not part of the selection. The colors are distinct groups, but the balls chosen within each group are not ordered.

The letter (a) is fixed and (b) is excluded. Choose (1) from (c,d) and (4) from the remaining (9), so (\binom{2}{1}\binom{9}{4}=252).

Chemistry is fixed and biology is excluded. Choose the remaining (4) subjects from (9) and subtract (\binom{7}{2}) selections containing both mathematics and physics to get (105).

The (3) numbers are fixed and (2) are excluded. The remaining (5) numbers are chosen from (13), so (\binom{13}{5}=1287).

For a failed triangle (3) points are chosen from the (10) collinear points. So the number is (\binom{10}{3}).

After removing (5) broken chairs (10) good chairs remain. The ways to choose (8) are (\binom{10}{8}=45).

If all special students are included there are (\binom{11}{4}=330) ways and if all are excluded there are (\binom{11}{7}=330) ways. The total is (660).

You can choose (4), (5), (6), or (7) from the last (7) questions. The sum of all valid cases is (3850).

The number of seniors can be (0), (1), (2), or (3). The total is (\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072).

Choose (1) of the two fixed cards and (5) cards from the remaining (13). The ways are (\binom{2}{1}\binom{13}{5}=2574).

There are \(2^{16}\) subsets of 16 distinct coins. The numbers of selections with an even number and an odd number of coins are equal, so the number of even selections is \(\frac{2^{16}}{2}=2^{15}=32768\). This includes the one selection in which no coin is chosen. The value 32767 would result if the empty selection were excluded. Exam tip: for \(n\geq1\) distinct objects, the numbers of even-sized and odd-sized subsets are each \(2^{n-1}\).

The number of odd selections is (2^{15-1}=16384). Remember that even and odd selections are equal in such questions.

The captain is excluded and the vice-captain is fixed. The remaining (6) players are chosen from (12), so (\binom{12}{6}=924).

The number of special objects can be (0), (1), or (2). The total is (\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335).

The required point divides every valid route into two independent parts. In the first part, the path contains 3 right moves and 2 up moves, five moves in all. Their order can be chosen in 5C3 = 10 ways. From that point to the destination, the remaining moves are 7 − 3 = 4 right and 4 − 2 = 2 up, six moves in all. This second part can be arranged in 6C4 = 15 ways, equivalently 6C2. By the multiplication principle, each first-part route can be paired with each second-part route, giving 10 × 15 = 150. Therefore option A is correct; adding the counts would be inappropriate because both parts must occur together.

Let xi be the number of identical balls placed in box i. The conditions are x1 + x2 + x3 + x4 = 7 and xi ≥ 1 for every box. First place one ball in each of the four distinct boxes; this uses 4 balls and guarantees that no box is empty. Three identical balls remain to be distributed among four boxes, now allowing zero additional balls in a box. By stars and bars, the number of distributions is C(3 + 4 − 1, 4 − 1) = C(6,3) = 20. Therefore option D is correct. Treating the balls as distinct would greatly overcount, and allowing empty boxes would count a different problem.