# Permutation and Combination Examples

Here you will learn some permutation and combination examples for better understanding of permutation and combination concepts.

Example 1 : If all the letters of the word ‘RAPID’ are arranged in all possible manner as they are in a dictionary, then find the rank of the word ‘RAPID’.

Solution : First of all, arrange all letters of given word alphabetically : ‘ADIPR’

Total number of words starting with A _ _ _ _ = 4! = 24

Total number of words starting with D _ _ _ _ = 4! = 24

Total number of words starting with I _ _ _ _ = 4! = 24

Total number of words starting with P _ _ _ _ = 4! = 24

Total number of words starting with R A D _ _ = 2! = 2

Total number of words starting with R A I _ _ = 2! = 2

Total number of words starting with R A P D _ = 1 = 1

Total number of words starting with R A P I _ = 1 = 1

$$\therefore$$ Rank of the word RAPID = 24 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 102

Example 2 : Find the number of ways of dividing 52 cards among 4 players equally such that each gets exactly one Ace.

Solution : Total number of ways of dividing 48 cards(Excluding 4 Aces) in 4 groups = $$48!\over (12!)^4 4!$$

Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = $$48!\over (12!)^4 4!$$ $$\times$$ 4!

Now, distribute these groups of cards among four players

= $$48!\over (12!)^4 4!$$ $$\times$$ 4!4! = = $$48!\over (12!)^4$$ $$\times$$ 4!

Example 3 : How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Solution : There are 4 odd digits (1, 1, 3, 3) and 4 odd places(first, third, fifth and seventh). At these places the odd digits can be arranged in $$4!\over 2!2!$$ = 6

Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in $$3!\over 2!$$ = 3 ways

$$\therefore$$     The required number of numbers = 6 $$\times$$ 3 = 18

Example 4 : In how many ways can 5 different mangoed, 4 different oranges & 3 different apples be distributed among 3 children such that each gets atleast one mango?

Solution : 5 different mangoes can be distributed by following ways among 3 children such that each gets at least 1 :

Total number of ways : ($$5!\over 3!1!1!2!$$ + $$5!\over 2!2!2!$$) $$\times$$ 3!

Now, the number of ways of distributing remaining fruits (i.e. 4 oranges + 3 apples) among 3 children = $$3^7$$ (as each fruit has 3 options).

$$\therefore$$     Total number of ways = ($$5!\over 3!2!$$ + $$5!\over {(2!)}^3$$) $$\times$$ 3! $$\times$$ $$3^7$$

Practice these given permutation and combination examples to test your knowledge on concepts of permutation and combination.