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6.Permutation and Combination
medium
A student is to answer $10$ out of $13$ questions in an examination such that he must choose at least $4$ from the first five question. The number of choices available to him is
A
$140$
B
$196$
C
$280$
D
$346$
(AIEEE-2003)
Solution
(b) As for given question two cases are possible.
$(i)$ Selecting $4$ out of first $5$ questions and $6$ out of remaining $8$ questions $ = {\,^5}{C_4}\, \times {\,^8}{C_6} = 140$ choices.
$(ii)$ Selecting $5$ out of first $5$ questions and $5$ out of remaining $8$ questions $ = {\,^5}{C_5}\, \times {\,^8}{C_5} = 56$ choices.
Total no. of choices = $140 + 56 = 196.$
Standard 11
Mathematics
