Practice Questions
Test your ability to select and combine items with these practical combination questions.
In how many ways can a team of 3 be chosen from 7 players?
Selection of 3 from 7 = ${}^7C_3 = (7 \times 6 \times 5) / (3 \times 2 \times 1) = 35$.
How many ways can 4 members be selected from 6 men and 4 women so that exactly two women are in the group?
Selection must have 2 Women AND 2 Men. ${}^4C_2 \times {}^6C_2 = 6 \times 15 = 90$.
The value of ${}^{10}C_8$ is equivalent to:
By symmetry, ${}^nC_r = {}^nC_{n-r}$, so ${}^{10}C_8 = {}^{10}C_2 = (10 \times 9) / 2 = 45$.
A box contains 5 red and 3 blue balls. In how many ways can 2 balls be selected such that at least one is blue?
Option 1: 1 Blue, 1 Red (${}^3C_1 \times {}^5C_1 = 3 \times 5 = 15$). Option 2: 2 Blue (${}^3C_2 = 3$). Total = $15 + 3 = 18$.
Evaluate ${}^{12}C_4$.
${}^{12}C_4 = (12\times11\times10\times9)/(4\times3\times2\times1) = 495$.
In how many ways can a team of 3 boys be selected from a group of 7 boys?
${}^7C_3 = (7\times6\times5)/(3\times2\times1) = 35$.
A committee of 6 is to be formed from 8 men and 4 women such that at least 2 women are included. In how many ways can this be done?
Cases: 2W+4M (${}^4C_2\times{}^8C_4=6\times70=420$), 3W+3M (${}^4C_3\times{}^8C_3=4\times56=224$), 4W+2M (${}^4C_4\times{}^8C_2=1\times28=28$). Total $=672$.
If ${}^nC_3 = 35$, find the value of $n$.
${}^nC_3 = n(n-1)(n-2)/6 = 35 \Rightarrow n(n-1)(n-2) = 210$. $n=7$ gives $7\times6\times5=210$.
In how many ways can 5 questions be selected from 10 questions if at least 2 out of 4 difficult questions must be included?
Cases: 2 difficult (${}^4C_2\times{}^6C_3=6\times20=120$), 3 difficult (${}^4C_3\times{}^6C_2=4\times15=60$), 4 difficult (${}^4C_4\times{}^6C_1=1\times6=6$). Total $=120+60+6=186$.
Apply symmetry to simplify ${}^{15}C_{13}$.
By symmetry ${}^nC_r = {}^nC_{n-r}$, so ${}^{15}C_{13} = {}^{15}C_2 = (15\times14)/2 = 105$.
Premium Content
Unlock Combination Quiz and all premium lessons with a subscription.
From ₹199.99/year — See plans