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The permutation is a way of filtering and selecting a set of objects, where the arrangement of objects does matter. However, the arrangement of objects may be done by imposing certain restrictions in the order of selection. For instance, the order of arrangement of articles, such that an article is always included or excluded from the set of given objects. Imposing the restrictions implies that not all the objects from the given set need to be ordered.
There are different types of common restrictions that may be imposed on the permutation:
Common types of restricted permutations are:
Also Read: Permutation and Combinations
r × n-1Pr-1
n-1Pr
Question 1. Find out how many 4 digits numbers without any repetition can be made using 1, 2, 3, 4, 5, 6, 7 if 4 will always be there in the number?
Solution:
Here to find 4 digits number without any repetition can be made using 1, 2, 3, 4, 5, 6, 7 if 4 will always be there in the number,
We will use the formula for
Number of permutations of 'n' things taken 'r' at a time. In which a particular thing always occur = r × n-1Pr-1Here,
- r = 4
- n = 7
Further putting values in the above formula
⇒ r × n-1Pr-1
⇒ 4 × 7-1P4-1
⇒ 4 × 6P3
⇒ 4 × 6!/3!
⇒ 4 × (6 × 5 × 4 × 3!)/3!
⇒ 480Therefore, 480 numbers can be made
Question 2. How many 5 digit numbers can be formed by 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. So that 2 is always there in the number?
Solution:
Here to find 5 digit numbers can be formed by 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. So that 2 is always there in the number,
We will use the formula for
Number of permutations of 'n' things taken 'r' at a time. In which a particular thing always occur = r × n-1Pr-1Here,
- r = 5
- n = 10
Further putting values in the above formula
⇒ r × n-1Pr-1
⇒ 5 × 10-1P5-1
⇒ 5 × 9P
⇒ 5 × 9!/(9-4)!
⇒ 5 × (9 × 8 × 7 × 6 × 5!)/5!
⇒ 15120Therefore, 15120 numbers can be made
Question 3. How many different three-letter words can be made by 5 vowels if 'a' is never included?
Solution:
Here to find different three-letter words can be made by 5 vowels if 'a' is never included,
We will use the formula for Number of permutations of 'n' things taken 'r' at a time. In which a particular thing never occurredn-1Pr
Here,
- r = 3
- n = 5
Further putting values in the above formula
⇒ n-1Pr
⇒ 5-1P3
⇒ 4P3
⇒ 4!/(4 - 3)!
⇒ 4 × 3 × 2
⇒ 24Therefore, 24 words can be made
Question 4. How many four-digit numbers without any repetition can be made by using 1, 2, 3, 4, 5, 6, 7 if 4 will never be included?
Solution:
Here to find four-digit numbers without any repetition can be made by using 1, 2, 3, 4, 5, 6, 7 if 4 will never be included,
We will use the formula for Number of permutations of 'n' things taken 'r' at a time. In which a particular thing never occurred
n-1Pr
Here,
- r = 4
- n = 7
Further putting values in the above formula
⇒ n-1Pr
⇒ 7-1P4
⇒ 6P4
⇒ 6!/(6 - 4)!
⇒ 6 × 5 × 4 × 3 × 2!/2!
⇒ 360Therefore, 360 numbers can be made