IIT JEE , AIEEE Forum - Mathematics , Algebra

anupamh.chakraborty

number of ways of selection of 8 letters from 24 letters of which 8 are a, 8 are b,and rest are unlike, is given by

a)2^7 b)8*2^8 c)10*2^7 d)none

malay

the answer is (c)

edison

Well the answer is (b)

The explanation is as follows:

Number of ways of selection of 8 letters from 24 letters of which 8 are a, 8 are b, is given by following cases

1) 8a = 1

2) (7a,1b) ; (7a,1 letter other than a and b) = 1 + ⁸C₁ = 1+8 = 9

3) (6a,2b); (6a,1b,1 other than a and b); (6a, 2 other than a and b) = 1+⁸C₁+⁸C₂

= 9 + 28 = 37

4) (5a,3b) and so on = 1+⁸C₁+⁸C₂+ ⁸C₃= 37 + 56 =93

5) (4a,4b)...and so on = 1+⁸C₁+⁸C₂+ ⁸C₃+⁸C₄= 93+70 = 163

6) (3a,5b)...and so on = 1+⁸C₁+⁸C₂+ ⁸C₃+⁸C₄1+⁸C₅= 163+56 = 219

7) (2a,6b)...and so on= 1+⁸C₁+⁸C₂+ ⁸C₃+⁸C₄1+⁸C₅1+⁸C₆=219+28 = 247

8) (a,7b)....so on = 1+⁸C₁+⁸C₂+ ⁸C₃+⁸C₄1+⁸C₅1+⁸C₆+⁸C₇=247+8 = 255

Similarly we can interchange the role of 'a' and 'b' to get exactly eight more cases as above.

so total number of possible selections are = 2(1+9+37+93+163+219+247+255)

= 2 x 1024 = 2¹¹ = 8 x 2⁸