 # Permutations with Restrictions

Suppose a class teacher of a 25 students wants to select three students for three different posts of responsibility. In how many ways can she do so? Being familiar with the concepts of permutations you can answer it. Clearly, there are 25P3 = 25! ⁄ 22! ways in which the teacher can choose any three students. But what if she puts on some restrictions like for a certain post, a student must have scored above 85% in a certain test and likewise other similar restrictions. How can you deal with such types of permutations? Let us get familiar with the concept of permutation with restrictions.

### Suggested Videos        ## Permutations with Restrictions

A permutation is an arrangement of a set of objects in an ordered way. An addition of some restrictions gives rise to a situation of permutations with restrictions. Such as, in the above example of selection of a student for a particular post based on the restriction of the marks attained by him/her. Obviously, the number of ways of selecting the students reduces with an increase in the number of restrictions.

The most common types of restrictions are that we can include or exclude only a small number of objects. In other words, a certain set of objects will either come together or always stay apart. This means that not all the objects need to be ordered.

#### The common types of restricted permutations are:

• Formation of numbers with digits with some digits at fixed positions.
• A set of objects either always occur or never occur.
• Word building with some letters with a fixed position.
• Vowels or consonant occur together.
• Restrictions for circular permutations.

In our daily lives, we can find a lot of examples of permutation with restrictions like the decision for the order of eating, the choice of dress to wear, the combinations of the colours to make etc.

## Basic Rules in Permutations

• Suppose we have n letters or items out of which t are of the same kind and the rest are all different = n! ⁄ t!.
• Number of permutation of n items, taken t at a time, when we include a particular item in each arrangement is n – 1 P t – 1 × t
• When a particular thing is fixed, number of permutation of n items out of which t no. of items are taken at a time = n−1Pt−1
• The number of permutations of n items, taken t at a time when a particular thing is never taken = n−1 t
• The number of permutations for n items, taken t at a time when p specified things always come together = n! × (n – p + 1)!.

The restriction is also applicable to the circular permutations.

## Solved Examples for You

Problem: Consider a word ‘YOURSELEVES’. In how many ways the letter can be arranged if U and S always come together and ‘U’ always precedes ‘S’?

Solution: The word ‘YOURSELVES’ has 11 letters out of which ‘S’ repeats two times and ‘E’ repeats three times. The rest are all different.

If the letter ‘U’ and ‘S’ come together, they are considered as one letter. The remaining 10 letters can rearrange themselves in 10! ⁄ (2! 3!) =  302400 ways.

### Problem:

Find the number of ways in which five persons A, B, C, D, and E sit around a round table such that

1. There is no restriction.
2. A and D must always sit together.
3. C and E must not sit together.
1. Five persons can sit around a roundtable in (5 – 1)! = 4! = 24 ways.
2. Since A and D sit together in all the possible arrangement, we have to consider them as one unit. Now, the rearrangement to be done is for four people only. The number of ways of arrangement in a circular permutation is (4 – 1)! = 3! = 6. A and D can interchange their positions in 2 ways. So, the required number of ways of rearrangement is 6 × 2 = 12. 3. We have to find the number of ways in which C and E must not sit together. The number of ways in which the arrangement is possible is the same as the difference between the total number of ways and the number of the ways in which C and E can sit together.

From the second part of the question, it is clear that the number of ways in which the persons C and E can sit together is 12. From the first part of the question, we get the total number of ways of possible arrangements = 24. So, the required number of the ways in which C and E do not sit together = 24 – 12 = 12 ways.

Share with friends
Browse
##### Permutations and Combinations
Customize your course in 30 seconds

Which class are you in?

5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.  Ashhar Firdausi
IIT Roorkee
Biology  Dr. Nazma Shaik
VTU
Chemistry  Gaurav Tiwari
APJAKTU
Physics
Get Started
Browse
##### Permutations and Combinations
Customize your course in 30 seconds

Which class are you in?

No thanks.