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There are 9 students in the chess club. The club moderator selects 3 of them to go to a chess tournament. In how many ways can she choose 3 of the 9 students?A.126B.84C.36D.504

Question

There are 9 students in the chess club. The club moderator selects 3 of them to go to a chess tournament. In how many ways can she choose 3 of the 9 students?A.126B.84C.36D.504

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Solution

The problem can be solved using the combination formula which is used when the order of selection does not matter. The formula is:

C(n, r) = n! / [r!(n-r)!]

where:

  • n is the total number of items,
  • r is the number of items to choose,
  • "!" denotes factorial, which means multiplying all positive integers up to that number.

In this case, n = 9 (the total number of students) and r = 3 (the number of students to be selected for the tournament).

So, the calculation would be:

C(9, 3) = 9! / [3!(9-3)!] = (987654321) / [(321)(654321)] = (987) / (321) = 84

So, the club moderator can choose 3 students out of 9 in 84 different ways. Therefore, the correct answer is B.84.

This problem has been solved

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