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Solutions

  1. Answer: \binom{10}{3} \cdot \binom{25}{4} = 1,518,000

  2. 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})

  3. Answer: \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} = 120 + 45 +10 + 1 = 176

  4. 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 ).

  5. Answer:
    1. C(52, 5) = 2, 598, 960
    2. \binom{52}{5} \cdot  \binom{47}{5} \cdot \binom{42}{5} \cdot \binom{37}{5}

  6. Answer: \binom{4}{2} \cdot \binom{48}{3} = 6 \cdot 17296 = 103776

  7. Answer: \binom{12}{3} \cdot \binom{9}{4} \cdot \binom{5}{5}

  8. Answer:
    1. \binom{10}{2} = 45
    2. \binom{10}{3} = 120

  9. 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 |}

  10. 17. Hint: 9998 = 10000 − 2

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