Question

A committee of 11 members is to be formed from 8 males and 5 females. Let m denotes the number of ways of selecting committee having atleast 6 males and n denotes the number of ways of selecting atleast 3 females then

• A. m = n - 8
• B. m = n = 68
• C. m = n = 78
• D. m = n = 65

Answer 1

Answer: (C)

m = n = 78

Answer 2

Given that: ($8$ males, $5$ females)
Committee to be selected = $11$ members
$m$ = no. of ways the committee is formed with at least 6 males.
$(6M, 5F)$ or $(7M, 4F)$ or $(8M, 3F)$
$= ^8C_6 × ^5C_5+ ^8C_7 × ^5C_4+ ^8C_8 × ^5C_3$
$= 78$

$n$ = no. of ways the committee is formed with atleast 3 female
$(8M, 3F)$ or $(7M, 4F)$ or $(6M, SF)$
$= ^8C_8 × ^5C_3 + ^8C_7 × ^5C_4 + ^8C_6 × ^5C_5$
$= 10 + 40 + 28$
$= 78$
$⇒ m = n = 78$

So, the correct option is (C): $m = n = 78$