Tips And Tricks And Shortcuts For Permutation And Combination

In how many ways can a word be arranged.

Tricks and Tips on type 1 Question

This is Permutation Question.

Let us take this ahead as an example – 

In how many ways can the letters of the word ‘LEADER’ be arranged?

Count number of Occurances

• L – 1
• E – 2
• A – 1
• D – 1
• R – 1

Total Unique Occurrences – 6(as E repeated 2 times)

Direct Formula = (Unique Occurrences)!/(Each Individual Unique Occurrences)

so = 6!/(1!)(2!)(1!)(1!)(1!) = 360

In how many ways x objects out of a and y objects out of b can be arranged.

Tips and Tricks type 2 problems

Let us take this as well with an example –

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (7C3 x 4C2)

=		7 x 6 x 5	x	4 x 3
		3 x 2 x 1		2 x 1

= 210.Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.

Required number of ways = (210 x 120) = 25200.

There are x objects and y objects, a from x has be selected and b from y. How many ways can it be done when N Number of objects from x should always be selected

Tricks and Tips Type 3 Problems

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways	= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
	= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
	= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5 2 x 1 2 x 1 3 x 2 x 1 2 x 1
	= (24 + 90 + 80 + 15)
	= 209.

Coloured Ball Questions

Tricks and Tips Type 4 Problems

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
	=   3 x 6 x 5 + 3 x 2 x 6 + 1 2 x 1 2 x 1
	= (45 + 18 + 1)
	= 64.

Circular Combinations Problems

If 6 people are going to sitting at a round table, but Sam will not sit next to Suzie, how many different ways can the group of 6 sit?

Couple of of ways of doing this: 

First: 
a. Total circular permutations = (6-1)! = 5! = 120. 
b. Ways in which Sam and Suzie sit together = 2! * 4! = 2*24 = 48 
Required ways = Total – Together = 120 – 48 = 72. 

Second: 
a. We have total of 6 places. Fix Suzie. Now Sam can’t sit at either seat beside her. So number of places where Sam can sit = 5-2 = 3. 
For the other 4 people we can arrange them in 4! ways in 4 seats. 
So total ways = 3 * 4! = 72.