Permutations Concept and Questions:Crack CAT Quantitative Aptitude

Feb 2, 2013, 11:30 IST

Permutations and Combinations is a very important chapter for CAT Entrance.These questions demand ability to comprehend and interpret the problem situation and then find an appropriate way to arrive at the answer. This will be amply evident through the examples that we undertake in this topic.

These questions demand ability to comprehend and interpret the problem situation and then find an appropriate way to arrive at the answer. This will be amply evident through the examples that we undertake here.

Permutations

Suppose we have n distinct objects. Each ordered arrangement of these objects is called permutation of these n objects.The total number of different permutation of n objects is:

     nx (n-1) x (n-2) x...x 3 x 2 x 1 = n!

Example:

How many different batting orders are possible for a cricket team?

Solution:

Since a cricket team consists of 11 players, there are

11! = 11 x 10 x 9 x...x 2 x 1 = 39,916,800 possible batting orders.

Example:

Fatima has 4 management books, 3 mathematics books, 2 history books and 2 geography books. How many different arrangements of books are possible on her bookshelf, if all the books on the same subject must be placed together?

Solution:
If she first puts the management books, then the mathematics books, then the history books, and finally the geography books, the number of possible arrangements is:

(4! x 3! x 2! x 2!)= 576

There are 4! possible ways in which the four subjects can be arranged. Hence the total number of arrangements of books on her bookshelf is:

4! x (4! x 3! x 2! x 2!) = 13,824

Permutations when some of the Object are Alike
The number of different permutations of n objects of which n1 are alike n2 are alike… nkare alike is:   (the proof of this is on the lines of the steps shown in the example below)

Example 7:

How many different five-letter words can be formed using the letter from the world APPLE?

Solution:

If the two P’s were distinct (they could have different subscripts and colours), the number of possible permutations would have been 5! = 120.

For example let us consider one permutation: P1LEAP2

Now if we permute the P’s amongst them we still get the same word PLEAP. The two P’s can be permuted amongst them in 2! ways.

We were counting P1LEAP2 and P2LEAP1 asdifferent arrangements only because we were artificially distinguishing between the two P’s Hence the number of different five letter words that can be formed is

Permutations and Combinations is also about logical interpretation and analytical skills.

Jagran Josh
Jagran Josh

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