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

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 ⁿP_r, in which;

ⁿP_r = 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:

ⁿP_r = 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,

ⁿP_r = 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,

ⁿP_r = 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

ⁿP_r = 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:

ⁿP_r = 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.