Question 7 and 8, Exercise 6.3

Solutions of Question 7 and 8 of Exercise 6.3 of Unit 06: Permutation and Combination. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

A committee of $5$ members is to be formed out of $6$ men and $4$ women. In how many ways can it be done if it has exactly $2$ women.

Solution.

(i) If there are exactly $2$ women then there will be $3$ man in committee.

Total possible ways $={ }^{4} C_{2} \times{ }^{6} C_{3}=6 \times 20=120$

A committee of $5$ members is to be formed out of $6$ men and $4$ women. In how many ways can it be done if it has at least $2$ women.

Solution.

At least $2$ women means there could be more than $2$ women as well.
So we will calculate possibilities of having $2,3$ and four women in committee.
So if there are $2$ women we already had calculated $120$ possibilities.
If there are $3$ women then $2$ men are in committee.
Then possible ways are $={ }^{4} C_{3} \cdot{ }^{6} C_{2}=4 \times 15=60$
If $4$ women are in committee then one man is there in committee.
so total possible ways are $={ }^{4} C_{4} \times{ }^{6} C_{1}=1 \times 6=6$
Hence total possible ways if at least two women are in committee are $=120+60+6=186$

A committee of $5$ members is to be formed out of $6$ men and $4$ women. In how many ways can it be done if it has at most $2$ women?

Solution.

At most to women mean either $1$ or $2$ women could be selected in committee.
If one women is selected than $4$ men would be seitcted,
and possible ways $={ }^{4} C_{1} \times{ }^{6} C_{4}=4 \times 15=60$
If $2$ women are selected then $3$ men are there is committee.
So possible ways $={ }^{4} C_{2} \times{ }^{6} C_{3}=6 \times 20=120$
If no woman selected $={ }^{6} C_{5}=6$
Total possible ways for at most two women
$$ =60+120+6=186 $$

There are $10$ points on circle. Find the number of lines?

Solution.

For a line, we need only two points so number of ways to choose $2$ points out of $10$ are $={ }^{10} C_{2}=45$ ( (ii) For triangle we need 3 points and number of ways to choose 3 points out of 10 are $={ }^{10} C_{3}=120$

There are $10$ points on circle. Find the number of triangles that can be drawn?

Solution.

For triangle we need $3$ points and number of ways to choose $3$ points out of $10$ are $={ }^{10} C_{3}=120$

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