Question

A question paper has two parts P and Q each containing $ 10 $ questions. If a student needs to choose $ 8 $ from part A and $ 4 $ from part B, in how many ways can he do that?
A. $ 9450 $
B. $ 2524 $
C. $ 5251 $
D.None of these

Answer 1

Hint: Use the combinations formulas to find the combination of $ 8 $ question from part A out of total $ 10 $ questions and $ 4 $ questions from part B out of the total $ 10 $ questions and then multiply those combinations to get the total number of combinations.

Complete step-by-step answer:
A question paper has two parts P and Q each containing $ 10 $ questions. A student needs to choose $ 8 $ from part A and $ 4 $ from part B.
As it is given in the question that the total number of questions in part A are $ 10 $ and the student needs to select $ 8 $ questions from it. So, the possible number of combinations for it is equal to $ ^{10}{C_8} $ .
Also it is given in the question that the total number of questions in part B is $ 10 $ and the student needs to select $ 4 $ questions from it. So, the possible number of combinations for it is equal to $ ^{10}{C_4} $ .
So, the total number of combinations for the whole paper is equal to $ ^{10}{C_8}{ \times ^{10}}{C_4} $ .
Evaluate the number of ways of selecting all the questions in the exam:
 $
\Rightarrow ^{10}{C_8}{ \times ^{10}}{C_4} = \dfrac{{10!}}{{8!\left( {10 - 8} \right)!}} \times \dfrac{{10!}}{{4!\left( {10 - 4} \right)!}} \\
   = \dfrac{{10!}}{{8!2!}} \times \dfrac{{10!}}{{4!6!}} \\
   = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1 \times 8!}} \times \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{4 \times 3 \times 2 \times 1 \times 6!}} \\
   = 45 \times 210 \\
   = 9450 \;
  $
So, the number of ways to set the questions is equal to $ 9450 $ .
So, the correct answer is “ $ 9450 $ ”.

Note: The number of combinations for selecting $ r $ items out of the total items $ n $ is equal to $ ^n{C_r} $ . The formula for evaluating the combinations is given as $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ .