AMC 8 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 12 Medium

3 of the 5 children are selected to participate in a competition, and Chris and Debbie are 2 of the 5 children. If at least one people between Chris and Dibbie is selected, how many ways of selecting participants are there?

  • A.

    17

  • B.

    2021

  • C.

    9

  • D.

    5

  • E.

    1

Answer:C

From 5 children, choosing 3 can be done in _{5}C_{5} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 ways.

The case where neither Chris nor Debbie is chosen means choosing all 3 from the remaining 3 children, which is _{3}C_{3} = 1 way.

Therefore, the number of ways in which at least one of the two is chosen is 10 - 1 = 9.

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