Combinations

By Pascal's identity (\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}). Hence the answer is (\binom{11}{5}).

Using Pascal’s identity, \(\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}\), we get \(\binom{12}{5}=\binom{11}{5}+\binom{11}{4}\). Hence, \(\binom{12}{5}-\binom{11}{5}=\binom{11}{4}=330\). The value 462 is \(\binom{11}{5}\), not the required difference. Exam tip: for differences of adjacent binomial coefficients, apply Pascal’s identity to identify the remaining term directly.

The element (1) is already chosen so the remaining (4) elements are chosen from (9). The number of ways is (\binom{9}{4}=126).

Total subsets are (\binom{10}{5}=252) and those containing both (2), (3) are (\binom{8}{3}=56). Hence (252-56=196).

The even numbers from 1 to 22 are 2, 4, 6, ..., 22, so there are 11 even numbers. Since the order of selection does not matter, use combinations: \(\binom{11}{4}=330\). Therefore, the correct answer is 330. The value 462 equals \(\binom{11}{5}\), so it does not represent choosing 4 numbers. Exam tip: first count the eligible numbers, then apply \(\binom{n}{r}\) for the required selection.

The relevant principle is to count the odd and even selections separately and then multiply them. From 1 through 24 there are 12 odd numbers and 12 even numbers. The number of ways to choose 3 odd numbers is C(12,3) = 12 × 11 × 10 ÷ 6 = 220. The number of ways to choose 2 even numbers is C(12,2) = 12 × 11 ÷ 2 = 66. Because a valid selection requires both an odd group and an even group, the multiplication rule gives 220 × 66 = 14,520. Hence option C is correct. The other choices do not equal the product of the two required combinations and therefore omit or miscount some selections.

After removing (5) special students (11) students remain. Hence there are (\binom{11}{6}=462) ways.

Total ways are (\binom{18}{6}=18564). Removing the cases of (0) teacher and (1) teacher gives (18564-\binom{10}{6}-\binom{8}{1}\binom{10}{5}=16338).

Total ways are (\binom{15}{5}=3003). Removing all-white (\binom{7}{5}=21) and all-black (\binom{8}{5}=56) gives (2926).

Since the prizes are identical, the order of the 4 selected candidates does not matter. Therefore, the required number of combinations is
\(\binom{12}{4}=\frac{12!}{4!\,8!}=495\). Option 11880 is \(12P4\), which counts different orders separately, so it is not applicable here. Exam tip: use combinations when a selection is made and the objects are identical.

Total ways are (\binom{10}{5}=252) and if both special friends do not go then (\binom{8}{5}=56). Hence (252-56=196).

Choose (2) from the special fruits and (3) from the remaining (7). The ways are (\binom{4}{2}\binom{7}{3}=210).

The balls are separated into three colors, and the required number must be chosen independently from each color. Since the balls are being selected, not arranged, order does not matter within any color. Thus combinations are used. The choices made from the red, blue, and green groups are independent, so the multiplication principle tells us to multiply the three combination counts.

Choose one red ball in \(\binom{6}{1}=6\) ways, two blue balls in \(\binom{7}{2}=21\) ways, and two green balls in \(\binom{8}{2}=28\) ways. Therefore the total is \(6\times21\times28=3528\). Hence option B is correct. Adding the counts would be wrong because one complete selection includes choices from all three colors simultaneously.

The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (7). The ways are (\binom{7}{3}=35).

Total ways are (\binom{8}{4}=70) and if both special subjects are included there are (\binom{6}{2}=15) ways. Hence (70-15=55).

The (2) numbers are fixed and (2) are 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 (7) collinear points. So the number is (\binom{7}{3}).

After removing (3) broken chairs (7) good chairs remain. The ways to choose (5) are (\binom{7}{5}=21).

If both are included there are (\binom{9}{3}=84) ways and if both are excluded there are (\binom{9}{5}=126) ways. The total is (210).

The cases are choosing (2), (3), or (4) from the last (4). The total is (\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185).

The number of seniors can be (0), (1), or (2). The total is (\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772).

Choose (1) of the two fixed cards and (4) cards from the remaining (10). The ways are (\binom{2}{1}\binom{10}{4}=420).

The direct answer is option B, 512. For each of 10 different coins, there are two choices: select it or do not select it, giving \(2^{10}\) total subsets. Exactly half of all subsets have an even number of coins and half have an odd number, because pairing a selection with one obtained by changing one fixed coin switches parity. Therefore the number of even selections is \(2^{10}/2=2^9=512\). This includes selecting zero coins and all 10 coins. Option A, 256, is only half of the correct even count. Option B is correct. Option C, 1023, counts all nonempty selections, not only even ones. Option D, 1024, counts every selection, including odd selections. Exam cue: for \(n\) distinct objects, the number of subsets of even size is \(2^{n-1}\).

There are \(2^9=512\) subsets of 9 distinct toys. Pair every selection not containing one fixed toy with the selection obtained by adding that toy. Each pair has one even-sized selection and one odd-sized selection. Hence, the number of odd-sized selections is half of 512: \(2^9/2=2^8=256\). Thus, 255 is incorrect because the numbers of even-sized and odd-sized selections are equal. Exam tip: For \(n\geq1\) distinct objects, the number of odd-sized subsets is \(2^{n-1}\).

The fixed captain is not to be included in the selection, so only 9 players are available. The number of ways to select 4 of them is
\(\binom{9}{4}=
\frac{9!}{4!5!}=126
\). The value 210 is
\(\binom{10}{4}\), which would result if the captain were not excluded. Exam tip: First remove any person who is explicitly excluded from the total number of choices.