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2025 National One Eighth mathematics Topic 15 Free

Permutations and combinations

Selecting committees of men and women using combinations, e.g. $\binom{m}{k}\binom{w}{l}$ · Sub-topic 1

ONE EIGHTH CONTEST 2025

St. Peter's SHS - 38 points

Osei Kyeretwie SHS - 38 points

Okuapemman School - 37 points


ONE EIGHTH CONTEST 2025

University Practice SHS - 43 points

Ghana SHS, Koforidua - 38 points

Winneba SHS - 30 points


ONE EIGHTH CONTEST 2025

Mawuli School - 48 points

Armed Forces SHTS - 38 points

Berekum Presby SHS - 33 points


ROUND 5

TIE BREAKER

TIE BREAKER

QUESTION

Find the number of ways of selecting 6 men and 6 women from a group of 8 men and 7 women

SOLUTION

$={}_{6}^{8}C\times{}_{6}^{7}C$

$=28\times 7$

$=196$


PRACTICE QUESTIONS


1. In how many ways can a committee of 3 men and 2 women be selected from 5 men and 4 women?

ANSWER: 60


SOLUTION

$\text{Ways to choose the men}=\binom{5}{3}=10$

$\text{Ways to choose the women}=\binom{4}{2}=6$

Multiply the choices:

$10\times6=60$


2. In how many ways can a committee of 2 men and 3 women be selected from 6 men and 5 women?

ANSWER: 150


SOLUTION

$\text{Ways to choose the men}=\binom{6}{2}=15$

$\text{Ways to choose the women}=\binom{5}{3}=10$

Multiply the choices:

$15\times10=150$


3. In how many ways can a committee of 4 men and 1 women be selected from 7 men and 3 women?

ANSWER: 105


SOLUTION

$\text{Ways to choose the men}=\binom{7}{4}=35$

$\text{Ways to choose the women}=\binom{3}{1}=3$

Multiply the choices:

$35\times3=105$


4. In how many ways can a committee of 3 men and 3 women be selected from 8 men and 6 women?

ANSWER: 1120


SOLUTION

$\text{Ways to choose the men}=\binom{8}{3}=56$

$\text{Ways to choose the women}=\binom{6}{3}=20$

Multiply the choices:

$56\times20=1120$


5. In how many ways can a committee of 2 men and 2 women be selected from 4 men and 4 women?

ANSWER: 36


SOLUTION

$\text{Ways to choose the men}=\binom{4}{2}=6$

$\text{Ways to choose the women}=\binom{4}{2}=6$

Multiply the choices:

$6\times6=36$


6. In how many ways can a committee of 4 men and 2 women be selected from 9 men and 5 women?

ANSWER: 1260


SOLUTION

$\text{Ways to choose the men}=\binom{9}{4}=126$

$\text{Ways to choose the women}=\binom{5}{2}=10$

Multiply the choices:

$126\times10=1260$