# 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 ^{n}P_{r}, in which;

^{n}P_{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;

^{23}P_{5}= 23!/(23 – 5)!

^{23}P_{5 }= 23!/18!Simplifying

^{23}P_{5}= 23 × 22 × 21 × 20 × 19 × 18!/18!

^{23}P_{5}= 23 × 22 × 21 × 20 × 19

^{23}P_{5}= 4037880Therefore,

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:**

^{n}P_{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;

^{20}P_{3}= 20!/(20 – 3)!

^{20}P_{3}= 20!/17!Simplifying

^{20}P_{3}= 20 × 19 × 18 × 17!/17!

^{20}P_{3}= 20 × 19 × 18

^{20}P_{3}= 6840Therefore,

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,

^{n}P_{r}= n!/(n-r)!

^{4}P_{3}= 4!/(4-3)!

^{4}P_{3}= 4!/1!

^{4}P_{3}= 4 × 3 × 2 × 1

^{4}P_{3}= 24Hence,

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,

^{n}P_{r}= n!/(n-r)!

^{5}P_{4}= 5!/(5-4)!

^{5}P_{4}= 5!/1!

^{5}P_{4}= 5 × 4 × 3 × 2 × 1

^{5}P_{4}= 120Hence,

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

^{n}P_{r}= n!/(n-r)!

^{10}P_{4}= 10!/(10-4)!

^{10}P_{4}= 10!/6!

^{10}P_{4}= 10 × 9 × 8 × 7 × 6!/6!

^{10}P_{4}= 10 × 9 × 8 × 7

^{10}P_{4}= 5040In 5040 ways.

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

**Solution:**

^{n}P_{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;

^{16}P_{4}= 16!/(16 – 4)!

^{16}P_{4}= 16!/12!Simplifying

^{16}P_{4 }= 16 × 15 × 14 × 13 × 12!/12!

^{16}P_{4}= 16 × 15 × 14 × 13

^{16}P_{4}= 43680Therefore,

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