Combinations

By Pascal's identity (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the answer is (\binom{10}{5}).

Using Pascal’s identity, \(\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\). Therefore, \(\binom{11}{5}-\binom{10}{5}=\binom{10}{4}=210\). The value 252 is \(\binom{10}{5}\), not the required difference. Exam tip: apply Pascal’s identity directly when consecutive binomial coefficients are subtracted.

The element (1) is already chosen so the remaining (3) elements are chosen from (8). The number of ways is (\binom{8}{3}=56).

Total subsets are (\binom{9}{5}=126) and those containing both (2), (3) are (\binom{7}{3}=35). Hence (126-35=91).

The even numbers from 1 to 18 are 2, 4, 6, 8, 10, 12, 14, 16 and 18, so there are 9 even numbers. Since the order of selecting 4 numbers does not matter, use combinations: \(\binom{9}{4}=126\). Therefore, the correct answer is 126. The value 3024 is related to permutations, where the order of selection is also counted. Exam tip: When items are “chosen” and no order is specified, usually use \(nCr\).

The governing concept is the multiplication rule applied to independent combinations. The numbers from 1 to 20 contain 10 odd numbers and 10 even numbers. The number of ways to choose 2 odd numbers is C(10,2) = 10 × 9 ÷ 2 = 45. Independently, the number of ways to choose 2 even numbers is also C(10,2) = 45. Every valid selection consists of one odd-number selection together with one even-number selection, so the counts must be multiplied: 45 × 45 = 2025. Therefore, option A is correct. The other values arise from failing to multiply the two independent choices or from using an incorrect number of available numbers.

After removing (4) special students (11) students remain. Hence there are (\binom{11}{6}=462) ways.

Total ways are (\binom{16}{5}=4368). Removing the cases of (0) teacher and (1) teacher gives (4368-\binom{9}{5}-\binom{7}{1}\binom{9}{4}=3213).

Total ways are (\binom{13}{5}=1287). Removing all-white (\binom{6}{5}=6) and all-black (\binom{7}{5}=21) gives (1260).

Since the prizes are identical, the order of the 4 selected candidates does not matter. Therefore, use combinations: \(\binom{11}{4}=\frac{11\times10\times9\times8}{4\times3\times2\times1}=330\). Hence, the correct answer is 330. The value 7920 is \(^{11}P_4\), which would apply if the prizes were distinct. Exam tip: use combinations when identical prizes or objects are being selected.

Total ways are (\binom{9}{4}=126) and if both special friends do not go then (\binom{7}{4}=35). Hence (126-35=91).

Choose (2) from the special fruits and (3) from the remaining (6). The ways are (\binom{4}{2}\binom{6}{3}=120).

The choices for different colours are independent, so count each colour selection separately and multiply the results. From 5 red balls, one can be selected in \binom{5}{1}=5 ways. From 6 blue balls, two can be selected in \binom{6}{2}=15 ways. From 7 green balls, one can be selected in \binom{7}{1}=7 ways.

By the multiplication principle, the total is \binom{5}{1}\binom{6}{2}\binom{7}{1}=5\cdot15\cdot7=525. Therefore option C is correct. We do not multiply by an ordering factor because the balls are being selected, not arranged. The colour requirements are fixed, so each valid choice is counted exactly once.

The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (6). The ways are (\binom{6}{3}=20).

Total ways are (\binom{7}{4}=35) and if both special subjects are included there are (\binom{5}{2}=10) ways. Hence (35-10=25).

The (2) numbers are fixed and (1) is excluded. The remaining (3) numbers are chosen from (9) in (\binom{9}{3}=84) ways.

For a failed triangle (3) points are chosen from the (6) collinear points. So the number is (\binom{6}{3}).

After removing (3) broken chairs (6) good chairs remain. The ways to choose (5) are (\binom{6}{5}=6).

If both are included there are (\binom{8}{3}=56) ways and if both are excluded there are (\binom{8}{5}=56) ways. The total is (112).

The cases are choosing (2), (3), or (4) from the last (4). The total is (\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105).

The number of seniors can be (0), (1), or (2). The total is (\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316).

Choose (1) of the two fixed cards and (3) cards from the remaining (9). The ways are (\binom{2}{1}\binom{9}{3}=168).

The total number of subsets of 8 distinct coins is \(2^8=256\). The selections containing an even number of coins are \(\binom80+\binom82+\binom84+\binom86+\binom88\). The numbers of even-sized and odd-sized subsets are equal, so the number of even selections is \(2^8/2=2^7=128\). Option 127 omits the one way of selecting zero coins. Exam tip: Use \((1+1)^n\) and \((1-1)^n\) to show that the sums of even and odd binomial terms are equal.

The total number of subsets of 8 distinct toys is \(2^8=256\). By including or excluding one fixed toy, every even-sized selection can be paired with an odd-sized selection. Hence, the numbers of even-sized and odd-sized selections are equal, so the required number is \(256/2=128\). The value \(127\) is the number of non-empty even selections after excluding the empty selection, not the number of odd selections. Exam tip: For \(n\) distinct objects, the number of odd-sized selections is \(2^{n-1}\).

The captain is excluded from the selection, so only 8 players are available. The number of ways to choose 4 players from these 8 is

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