Kaplan GMAT Challenge Problem: Combinations

by on September 20th, 2010

We invite you to give our Combinations Challenge question below a try!  Aim for about 2 minutes and give it your best shot before reviewing the answer & explanation.  Good luck!

Question:

Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected.  How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?

A) 28
B) 46
C) 55
D) 63
E) 70

Solution:

The first step in this problem is to recognize that order does not matter.  We must create a committee of four people, but we are not putting these people in any sort of order.  Because order does not matter, we must use the combinations formula, which is n!/[k!(n-k)!], in order to determine the number of possible outcomes.

However, as is the case with most advanced combinations problems, the solution is not as simple as plugging a few numbers into the formula.  Instead, we need to consider exactly what types of outcomes we are looking for.

In this case, we need to select Jane and 3 other people, none of whom are Thomas; Thomas and 3 other people, none of whom are Jane; or Jane, Thomas and 2 other people.  Additionally, we need to note that this is an ‘or’ situation, which means that we need to add all of these possibilities together.

First, if we select Jane and three others, none of whom are Thomas, we must pick Jane plus 3 out of the remaining 6 people.  Because we can pick any 3 out of the 6, we set n = 6 and k = 3.  When we plug this into the formula it gives us 6!/(3!3!) = 20 (Quick review: 3! = 3 x 2 x 1, 6! = 6 x 5 x 4 x 3 x 2 x 1.)

Next we determine how many ways we can select three people, none of whom are Jane, to go along with Thomas.  This is identical mathematically, thus we end up with 20 in this case too.

Lastly, we determine how many outcomes are possible in which we have Jane, Thomas and 2 others.  We now select 2 out of 6, so n = 6 and k = 2.  6!/(4!2!) = 15.

All we need to do to get the answer is add up all of the possible outcomes.  20 + 20 + 15 = 55, which is the correct answer to this question (C).

8 comments

  • if the question above is changed to:

    what is the probability that both Jane and Thomas are selected in the group?

    Then the solution would be:

    15/70

    right?

    • Yes

    • How and Why?

  • required combinations = total combinations - combinations excluding both Jane and Thomas.

    Total combinations = 8c4 = 70

    combinations excluding both Jane and Thomas = 6c4 = 15

    Answer = 70-15 = 55

    • Wow!!!  Your approach was much simpler.  I must remember this.

  • I have a different approach to the question with the same outcome - just thought I would share it:

    Number of total different committees from the 8 people:
    8!/(4!*4!)= 70

    Number of outcomes without either Jane or Tomas:
    6!/(2!*4!)=15

    Number of total different committees from the 8 people - Number of outcomes without either Jane or Tomas:

    70-15=55

    Answer is C

  • I guess Gurpreet was faster ...

  • Does "if at least one of either Jane or Thomas is to be selected" not imply that Jane and Thomas could both could be in the same group?

    "if either Jane or Thomas is to be selected" would mean they cannot be in the same group.

    "atleast" means at the very least Jane or Thomas have to be selected but it does not exclude both being selected. Hence I think we should be picking 3 out of 7 and not 6.

    Can you please clarify

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