CAT 2025 Lesson : Permutations & Combinations - Binomial Expansion

In how many ways can $4$ students be selected from a group of $8$?

Solution

Note that here the arrangement does not matter. As it is only selections, we can apply the combinations formula.

Number of ways of selecting $4$ students from $8 =$ $^{8}C_{4} = \dfrac{8!}{4! \times (8 - 4)!} = \dfrac{8!}{4! \times 4!}$

$= \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 2 \times 7 \times 5 = 70$

Answer: $70$

In how many ways can $2$ or more people be selected from a group of $7$ people

Solution

We need to add all cases where exactly $2$ are selected, exactly $3$ are selected and so on till all $7$ are selected.

Number of ways in which $2$ or more are selected $=$ $^{7}C_{2}$ $+$ $^{7}C_{3}$ $+$ $^{7}C_{4}$ $+$ $^{7}C_{5}$ $+$ $^{7}C_{6}$ $+$ $^{7}C_{7}$

$= 2^{7} - (^{7}C_{0}$ $+$ $^{7}C_{1}) = 128 - (1 + 7) = \bm{120}$

Answer: $120$

Rasheed gets $7$ shots to fire at a target. He wins $1$ point if he hits the target and loses $1$ point if he misses the target. If the number of points Rasheed has at the end of $7$ shots is a positive number, then in how many different ways could he have managed this?

Solution

He gets positive points only if his hits are more than misses. Thus, his hits should have been $4, 5, 6$ or $7$.

Number of ways in which these papers can be selected $=$ $^{7}C_{4}$ $+$ $^{7}C_{5}$ $+$ $^{7}C_{6}$ $+$ $^{7}C_{7}$ $=$ $\dfrac{2^{7}}{2} = 2^{6} = 64$

Answer: $64$