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The organizer of a conference is selecting workshops to include. She will select from 9 workshops about chemistry and 8 workshops about biology. In how many ways can she select 8 workshops if 7 or more must be about chemistry?(If necessary, consult a list of formulas.)

Question

The organizer of a conference is selecting workshops to include. She will select from 9 workshops about chemistry and 8 workshops about biology. In how many ways can she select 8 workshops if 7 or more must be about chemistry?(If necessary, consult a list of formulas.)

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Solution

To solve this problem, we need to use the combination formula which is C(n, r) = n! / [r!(n-r)!] where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

There are two scenarios to consider:

  1. Selecting 7 workshops from chemistry and 1 from biology.
  2. Selecting all 8 workshops from chemistry.

Let's calculate each scenario:

  1. Selecting 7 from chemistry and 1 from biology: The number of ways to select 7 out of 9 chemistry workshops is C(9, 7) = 9! / [7!(9-7)!] = 36. The number of ways to select 1 out of 8 biology workshops is C(8, 1) = 8! / [1!(8-1)!] = 8. So, the total number of ways for this scenario is 36 * 8 = 288.

  2. Selecting all 8 from chemistry: The number of ways to select 8 out of 9 chemistry workshops is C(9, 8) = 9! / [8!(9-8)!] = 9.

So, the total number of ways to select 8 workshops if 7 or more must be about chemistry is 288 (from scenario 1) + 9 (from scenario 2) = 297.

This problem has been solved

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