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Numerical Ability
Permutation and Combination
A student is to answer 11 out of 14 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
A)198 B)264 C)342 D)228
Read Solution (Total 1)
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- 5c4*9c7+5c5*9c6
5c4=5
9c7=9*8/3
5c5=1
9c6=9*8*7/3*2
5+9*8/3+1*9*8*7/3*2=264 - 4 years agoHelpfull: Yes(3) No(0)
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