Class 11

Math

Algebra

Permutations and Combinations

An examination paper consists of $12$ questions divided into two parts $A$ & $B$. Part $A$, containing $7$ questions & part $B$ contains $5$ questions. A candidates is required to attempt $8$ questions selecting at least $3$ from each part. In how many ways can the candidates select the questions?

How can he go about (Set I, Set II) $=(3,5)$ or $(4,4)$ or $(5,3)$

So, total number of ways of doing this is given by

$5C_{3}×7C_{5}$ + $5C_{4}×7C_{4}$ + $5C_{5}×7C_{3}$

$=(10)(21)+(5)(35)+(1)(35)=210+210=420$