# Find the number of permutations of 5 CDs if you have a total of 23 CDs

• Last Updated : 29 Nov, 2021

Permutation is an ordered arrangement of all parts of a set of items. The sequence of occurrence of elements is taken into consideration in the case of permutations. It is an ordered combination.

For example, the number of permutations of n different objects taken r at a time is nPr, in which;

nPr = n!/(n-r)!

= n(n−1)(n−2)…{n−(r−1)}

(Note that n! = n×(n−1)×(n−2)×…×2×1)

### Find the number of permutations of 5 CDs if you have a total of 23 CDs.

Solution:

Given a total of 23 CDs, assuming that each CD is different, taking only 5 CDs at a time is given by;

23P5 = 23!/(23 – 5)!

23P5 = 23!/18!

Simplifying

23P5 = 23 × 22 × 21 × 20 × 19 × 18!/18!

23P5 = 23 × 22 × 21 × 20 × 19

23P5 = 4037880

Therefore,

The number of permutations of 5 CDs if we have a total of 23 CDs is 4037880.

### Sample Questions

Question 1.  Find the number of permutations of 3 students if you have a total of 20 students.

Solution:

nPr = n!/(n-r)!

Given a total of 20 students, assuming that each student is different, take only 3 students at a time is given by;

20P3 = 20!/(20 – 3)!

20P3 = 20!/17!

Simplifying

20P3 = 20 × 19 × 18 × 17!/17!

20P3 = 20 × 19 × 18

20P3 = 6840

Therefore,

The number of permutations of 3 students if we have a total of 20 students is 6840.

Question 2. Mallika has 4 chocolates and Mallika wants to give them to 3 beggars. Find out in how many possible ways can she do this?

Solution:

Here,

We have given that

n = 4

r = 3

Therefore,

nPr = n!/(n-r)!

4P3 = 4!/(4-3)!

4P3 = 4!/1!

4P3 = 4 × 3 × 2 × 1

4P3 = 24

Hence,

Mallika can give 4 chocolates to 3 beggars in 24 ways.

Question 3. Assume a set of 5 letters a, b, c, d, e.

Find out in how many ways we can select 4 letters without repetition?

Solution:

Here,

We have given that

n = 5

r = 4

Therefore,

nPr = n!/(n-r)!

5P4 = 5!/(5-4)!

5P4 = 5!/1!

5P4 = 5 ×  4 × 3 × 2 × 1

5P4 = 120

Hence,

We can select letters in 120 ways.

Question 4. Find out that how many four alphabets with or without meaning, can be made with the word, ‘LOGARITHMS’, Assume that the repetition of alphabets can not be done?

Solution:

Here

We have

n = 10

r = 4

nPr = n!/(n-r)!

10P4 = 10!/(10-4)!

10P4 = 10!/6!

10P4 = 10 × 9 × 8 × 7 × 6!/6!

10P4 = 10 × 9 × 8 × 7

10P4 = 5040

In 5040 ways.

Question 5. Find the number of permutations of 4 women if you have a total of 16 women.

Solution:

nPr = n!/(n-r)!

Given a total of 16 women, assuming that each student is different, take only 4 women at a time is given by;

16P4 = 16!/(16 – 4)!

16P4 = 16!/12!

Simplifying

16P4 = 16 × 15 × 14 × 13 × 12!/12!

16P4 = 16 × 15 × 14 × 13

16P4 = 43680

Therefore,

The number of permutations of 4 women if we have a total of 16 women is 43680.

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