ABC26GN5308 · Permutations and Combinations

Subject: General Aptitude · Chapter: Permutations and Combinations · Exam: 2005 · Marks: · Difficulty:

A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 woman. In how many different ways can it be done?

(a)364

(b)728

(c)931

(d)1001

(e)None of these

Answer

Answer (as printed): C

Explanation

Required number of ways = (6C1 × 8C3) + (6C2 × 8C2) + (6C3 × 8C1) + (6C4 × 8C0)  8 × 7 × 6  6 × 5 8 × 7 6 × + ×   3   2 × 1 2 × 1 = 6× 5× 4  + × 8 + (6 $\mathrm{C_{2}}$ × 1)  3   8 × 7 × 6 6× 5× 4  6× 5  = 6 ×  + 420 +  × 8 +  × 1  3 × 2 × 1  6  2×1  = (336 + 420 + 160 + 15) = 931.

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