# How many permutations can be formed by sampling 5 items out of 8?

• Last Updated : 23 Jun, 2022

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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