How many permutations can be formed by sampling 5 items out of 8?
Last Updated :
25 Dec, 2023
Permutation is defined as the process of structuring a group in which all of its members are organised in some sequence or order. It is a straightforward procedure of sequencing or organising each data or selected data from a batch of data or data distribution. It is one of the several ways of arranging items in a specific sequence. The sign nPr is used to denote the number of permutations of n unique objects taken r at a time.
Permutation formula
nPr = n! / (n – r)!
The above formula gives the number of permutations (arrangements) of n different objects if r objects are selected at a time and repetition is
not permitted. Here the value of r lies between 0 and n, such that 0 < r ≤ n.
If the repetition is allowed, the formula is given by the rth power of n, that is, nr.
Derivation
Suppose there are n objects out of which only r have to be selected at an instance. These objects further have to be filled in different containers every time r objects are selected.
Clearly, the number of permutations for each of these n distinct objects taking r entities at a time is nPr.
The number of ways first container can be filled is n ways.
The number of ways second container can be filled is (n – 1) ways.
The number of ways third container can be filled is (n – 2) ways.
The number of ways fourth container can be filled is (n – 3) ways.
Similarly for rth container the number of ways is (n – (r – 1)).
Now the total number of ways to fill r containers in continuation is the product of number of ways for all containers. This can be denoted by nPr as both have same meaning.
nPr = n (n – 1) (n – 2) (n – 3) . . . (n – (r – 1))
nPr = n (n – 1) (n – 2) … (n – r + 1)
Multiplying and dividing the RHS by (n – r) (n – r – 1) . . . 3 × 2 × 1 we have,
nPr = n! / (n – r)!
This derives the formula for number of permutations of n objects taken r at a time, such that 0 < r ≤ n.
How many permutations can be formed by sampling 5 items out of 8?
Solution:
We have, n = 8 and r = 5.
Case 1: If repetition is allowed
No of permutations = nr
= 85
= 8 × 8 × 8 × 8 × 8
= 32768
Case 2: If repetition is not allowed
Here, the number of permutations is given by,
8P5 = 8!/(8 – 5)!
= 8!/3!
= (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1)
= 8 × 7 × 6 × 5 × 4
= 6720
Similar Problems
Problem 1. How many permutations can be formed by sampling 4 of 10?
Solution:
We have, n = 10 and r = 4.
Case 1: If repetition is allowed
No of permutations = nr
= 104
= 10 × 10 × 10 × 10
= 10000
Case 2: If repetition is not allowed
Here, the number of permutations is given by,
10P4 = 10!/(10 – 4)!
= 10!/6!
= (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(6 × 5 × 4 × 3 × 2 × 1)
= 10 × 9 × 8 × 7
= 5040
Problem 2. How many permutations can be formed by sampling 3 of 7?
Solution:
We have, n = 7 and r = 3.
Case 1: If repetition is allowed
No of permutations = nr
= 73
= 7 × 7 × 7
= 343
Case 2: If repetition is not allowed
Here, the number of permutations is given by,
7P3 = 7!/(7 – 3)!
= 7!/4!
= (7 × 6 × 5 × 4 × 3 × 2 × 1)/(4 × 3 × 2 × 1)
= 7 × 6 × 5
= 210
Problem 3. Find the number of different ways in which the letters of the word “MATHS” can be formed if repetition is not allowed.
Solution:
The word “MATHS” has 5 letters. So, the value of n is 5.
Now the different ways in which the letters of the word “MATHS” can be formed is given by,
N = n!
= 5!
= 5 × 4 × 3 × 2 × 1
= 120
Problem 4. Calculate the unique 3 letter words that can be formed using the word “JACKPOT”.
Solution:
The word “JACKPOT” has 6 letters. So, the value of n is 6. Here, r = 3.
So, the number of permutations is given by,
6P3 = 6!/(6 – 3)!
= 6!/3!
= (6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1)
= 6 × 5 × 4
= 120
Problem 5. Calculate the unique 4 letter words that can be formed using the word “DEPOSIT”.
Solution:
The word “DEPOSIT” has 7 letters. So, the value of n is 7. Here, r = 4.
So, the number of permutations is given by,
7P4 = 7!/(7 – 4)!
= 7!/3!
= (7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1)
= 7 × 6 × 5 × 4
= 840
Problem 6. How many distinct 5 digit numbers can be formed from the number “12345678”.
Solution:
The number “12345678” has 8 digits. So, the value of n is 8. Here, r = 5.
So, the number of permutations is given by,
8P5 = 8!/(8 – 5)!
= 8!/3!
= (8× 7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1)
= 8 × 7 × 6 × 5 × 4
= 6720
Problem 7. Calculate the unique 5 letter words that can be formed using the word “BEAUTIFUL”.
Solution:
The word “BEAUTIFUL” has 9 letters. So, the value of n is 9. Here, r = 5.
So, the number of permutations is given by,
9P5 = 9!/(9 – 5)!
= 9!/4!
= (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(4 × 3 × 2 × 1)
= 9 × 8 × 7 × 6 × 5
= 15120
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