Permutation-Combination
Aptitude

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Q.

A student is required to answer 6 out of 10 questions divided into two groups each containing 5 questions.

He is not permitted to attempt more than 4 from each group. In how many ways can he make the choice?

 A.

210

 B.

150

 C.

100

 D.

200

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Solution:
Option(D) is correct

Number of ways of choosing 6 from 10,

$= {^{10}C_6}$

$=210$

Number of ways of attempting more than 4 from a group:

$=2×{^5C_5}×{^5C_1}$

$= 10$

⇒ Required number of ways,

$= 210 - 10$

$= \textbf{200}$


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