Try It Now

Work these exercises to see how well you understand this material.

Answer: $\binom{10}{3} \cdot \binom{25}{4} = 1,518,000$
Hint: Think of the set of positions that contain a 1 to turn this is into a question about sets.

Solution:
1. $\binom{8}{3}$
2. $2^8 - (\binom{8}{0} + \binom{8}{1})$
Answer: $\binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} = 120 + 45 +10 + 1 = 176$
Hint: Think of each path as a sequence of instructions to go right (R) and up (U).

Answer. Each path can be described as a sequence or R's and U's with exactly six of each. The six positions in which R's could be placed can be selected from the twelve positions in the sequence $\binom{12}{6}$ ways. We can generalize t his logic and see that there are $\binom{m + n}{n}$ paths from (0, 0) to (m, n ).
Answer:
1. C(52, 5) = 2, 598, 960
2. $\binom{52}{5} \cdot \binom{47}{5} \cdot \binom{42}{5} \cdot \binom{37}{5}$
Answer: $\binom{4}{2} \cdot \binom{48}{3} = 6 \cdot 17296 = 103776$
Answer: $\binom{12}{3} \cdot \binom{9}{4} \cdot \binom{5}{5}$
Answer:
1. $\binom{10}{2} = 45$
2. $\binom{10}{3} = 120$
Answer. Assume |A| = n. If we let x = y = 1 in the Binomial Theorem, we obtain $2^n = \binom{n}{0} + \binom{n}{1} + . . . + \binom{n}{n}$ , with the right side of the equality counting all subsets of A containing 0, 1 , 2, . . . , n elements. Hence $\left | P(A) \right | = 2^{\left | A \right |}$
17. Hint: 9998 = 10000 − 2

Answer: 10000³ − 3 · 2 · 10000² + 3 · 2² · 10000 − 2³ = 999, 400, 119, 992.