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JAMB 2014 Mathematics · Question 32

In how many ways can a team of 3 girls be selected from 7 girls?

Options

  1. A.7!3!\frac{7!}{3!}
  2. B.7!4!\frac{7!}{4!}
  3. C.7!3!4!\frac{7!}{3!4!}
  4. D.7!2!5!\frac{7!}{2!5!}

Answer: C. 7!3!4!\frac{7!}{3!4!}

Explanation written with AI and checked against the answer key

Analyze the question: We are choosing 3 girls from 7, and the order of selection doesn't matter. A team of Ada, Bola and Chika is the same team as Chika, Ada and Bola. So this is a combination problem.

Concept: The number of ways to choose r items from n items is

nCr=n!r! (n−r)!^nC_r = \frac{n!}{r!\,(n-r)!}

Solution:

Here n = 7 and r = 3, so

7C3=7!3! (7−3)!=7!3! 4!^7C_3 = \frac{7!}{3!\,(7-3)!} = \frac{7!}{3!\,4!}

Evaluating it:

7×6×53×2×1=2106=35\frac{7\times6\times5}{3\times2\times1} = \frac{210}{6} = 35

So there are 35 possible teams.

Checking the options:

  • A. 7!3!\frac{7!}{3!} = 840. It leaves out the (n−r)!(n-r)! part, so it is wrong.
  • B. 7!4!\frac{7!}{4!} = 210. This is the permutation 7P3^7P_3, where order matters, so it is wrong.
  • C. 7!3!4!\frac{7!}{3!4!} = 35. This is correct.
  • D. 7!2!5!\frac{7!}{2!5!} = 21. This is 7C2^7C_2, which is wrong.

Tip: If the question says "select", "choose" or "form a team/committee", order doesn't matter, so use combinations. If it says "arrange", "line up" or "positions", use permutations. In nCr^nC_r, the two numbers in the denominator (3 and 4) must add up to n (7). Option C passes this check and the others don't.

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