A student taking an examination is required to answer 10 out

A student taking an examination is required to answer 10 out of 15 questions. a) In how many ways can the 10 questions be selected? b) In how many ways can the 10 questions be selected if exactly 2 of the first 3 questions must be answered?

Solution

Here we have that r items out of total n items can be selected using the formula

nCr =n!/(r!(n-r)!)

Now for part (a)), we have n=15 and r=10, so total required ways

so 15C10 = 15!/(10!(15-10)!)

= (15 x 14 x 13 x 12 x 11 x 10!)/(10! x 5 x 4 x 3 x 2)

=(15 x 14 x 13 x 12 x 11 )/(15 x 4 x 2)

= 7 x 13 x 3 x 11 = 3003

this is the answer of part a>

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Now when 2 questions out of 3 will be selected in 3C2 ways and other 8 will be selected from remaining 12 questions in 12C8 ways.

So total ways = 3C2 x 12C8

=3!/(2!(3-2)! x 12!/(8!(12-8)!

=( 3 x 2!)/(2! x 1) x (12 x 11 x 10 x 9 x 8!)/ (8! x 4 x 3 x 2)

=(3 x 12 x 11 x 10 x 9)/( 4 x 6)

= 27 x 11 x 5

= 1484 ways.

Answer