GRE® Quant Practice | Permutation

GRE Sample Questions | Permutation Combination & Probability Questionbank

The GRE maths sample question given below is a problem solving question in permutation combination. This GRE sample question is an easy question.

Question 1: There are 5 doors to a lecture room. In how many ways can a student enter the room through a door and leave the room by a different door?

  1. 10
  2. 9
  3. 20
  4. 625
  5. 1024

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Video Explanation



Explanatory Answer

Step 1 of solving this GRE Permutation Question: Find the number of ways the student can enter the classroom

There are 5 doors to the classroom. The student can enter the class through any one of these doors.
It could be any one of A or B or C or D or E. So, there are 5 ways.

Step 2 of solving this GRE Permutation Question: Find the number of ways the student can leave the classroom through a different door.

The student cannot leave the classroom through the door she entered.
The number of choices to choose a door to leave is down to 4

Step 3 of solving this GRE Permutation Question: Compute the number of possibilities

  1. If the student entered through A, she could leave through B or C or D or E.
    There are 4 possibilities of entering through A and leaving through a different door viz., AB, AC, AD, AE.
  2. If the student entered through B, she could leave through A or C or D or E.
    There are 4 possibilities of entering through B and leaving through a different door viz., BA, BC, BD, BE.
  3. If the student entered through C, she could leave through A or B or D or E.
    There are 4 possibilities of entering through C and leaving through a different door viz., CA, CB, CD, CE.
  4. If the student entered through D, she could leave through A or B or C or E.
    There are 4 possibilities of entering through D and leaving through a different door viz., DA, DB, DC, DE.
  5. If the student entered through E, she could leave through A or B or C or D.
    There are 4 possibilities of entering through E and leaving through a different door viz., EA, EB, EC, ED.

There are a total of 4 + 4 + 4 + 4 + 4 = 20 ways.
We could also arrive at the same answer by saying - there are 5 ways to enter AND 4 ways to leave.
So, a total of 5 × 4 = 20 ways.

Alternative ways to express the answer

The answer to this question may also be expressed in alternative ways. Here are couple of those and the rational behind those expressions.
Ask yourself the following two questions

Q1. Sampling with replacement or without replacement?

Is the problem in hand one of sampling with replacement or sampling without replacement
If the outcome of the first sampling is not put back into the sample space, we are dealing with sampling without replacement.

An effective way of determining will be to look at the choices available for each of the sampling. There were 5 choices to select a door to enter. But we were left with only 4 to exit. So, the door selected to enter is no longer available in the sample space when we had to select a door to exit.

If we had 'n' choices for the first sampling and it reduces to (n -1) and further down to (n -2) and so on, we are dealing with sampling without replacement. So, without doubt this question is an example of "sampling without replacement".

If the question is a "sampling without replacement question", the number of ways of selecting r objects from n objects is nCr.
So your first step is to compute this value. For this question, we can select 2 doors out of 5 in 5C2 ways.

Q2. Whether or not order matters in this sampling?

One of the outcomes was selecting doors AB. Another one of the outcomes was selecting doors BA.
Are these two choices the same or are they different?

  • If AB is the same as BA, then order does not matter.
  • If AB is not the same as BA, then order matters.

In this question, entering by A and leaving by B is entirely different from entering by B and leaving by A. Therefore, AB is not the same as BA. So, order matters.

If order matters, multiply the answer you got in the previous step with the number of ways things can be reordered
r distinct objects can be reordered in r! ways.
Therefore, total outcomes = 5C2 * 2!

The same may also be expressed as 5P2
nPr = nCr * r!

Choice C is the correct answer

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