Restricted Permutations

Ans: Case 1: 2 black balls and four white balls

Case 2: 3 black balls and three white balls

Let the number of black balls in case 1 be b1 and the number of black balls in case 2 be b2.

Similarly, the number of white balls in case 1 be w1 and number of white balls in case 2 be w2.

Given, number of objects, n = 6

            b1 = 2

            w1 = 4

            b2 = 3

            w2 = 3

 Now,

= [n! / (b1! x w1!)] + [n! / (b2! x w2!)]

= [6! / (2! x 4!)] + [6! / (3! x 3!)]

= 15 + 20

= 35

Therefore, 35 different arrangements of the balls can be drawn.

Ans: a) In the word “SOUTHERN”, there are five consonants.

Hence, it can be arranged in 5! Ways.

= 5!

= 5 x 4 x 3 x 2 x 1

= 120

Therefore, the number of arrangements where the word contains only consonants are 120

1. b) It is given, the word should start with S and end with N

So, the word is going to look like this: S _ _ _ N

Since there are six letters between S and N in the word SOUTHERN, we have to arrange three letters from the set of 6 letters:

= nPr

= 6P3

= (6! / (6-3)!)

= [(6 x 5 x 4 x 3!) / 3!]

= 120

Therefore, there can be 120 arrangements where the word starts with S and ends with N.

1. c) It is given that the word should contain the letter U

So, the word is going to look like this: 

U _ _ _ _ or _U _ _ _ or _ _ U _ _ or _ _ _ U _ or _ _ _ _ U

i.e., five different places for U

Since four letters can fit when one of the letters of the 5-letter word is U,

Hence, we have to arrange four letters from 7 letters.

= nPr

= 7P4

And since there are five different places where U can be placed,

Hence, we multiply 7P4  with 5

= 7P4 x 5

= (7! / (7-4)!) x 5

= [(7 x 6 x 5 x 4 x 3!) / 3!] x 5

= 840 x 5

= 4200

Therefore, there can be 4200 arrangements where the word has the letter U.

1. d) It is given that the word should have T and H together

So, the word is going to look like this: 

T H _ _ _ or _T H _ _ or _ _ T H _ or _ _ _ T H   (also the positions of ‘T’ and ‘H’ can be interchanged)

i.e., four different places for T and H to occur together.

Since three letters can fit when the other two letters of the 5-letter word are T and H (which stays together),

Hence, we have to arrange three letters from the set of 6 letters.

= nPr

= 6P3

And since there are four different places where the letters T and H can be placed together,

Hence, we multiply 6P3 with 4

= 6P3 x 4

And since the places of T and H can also be changed between themselves

So, we multiply it by two as well

= 6P3 x 4 x 2

= (6! / (6-3)!) x 4 x 2

= [(6 x 5 x 4 x 3!) / 3!] x 4 x 2

= 120 x 4 x 2

= 960

Therefore, there can be 960 arrangements where the word has the letters T and H together in it.