a) A committee of five persons is to be formed from 5 men and 3 women. In how many ways can this be done so that at least two women are included? [4] OR Show that the number of combinations of 'n'…

a) Committee Formation with At Least Two Women

Given:

• Total men = 5
• Total women = 3
• Committee size = 5
• Condition: At least two women must be included.

Approach: We use combinations since the order of selection does not matter. The phrase "at least two women" means we consider cases where the committee has 2, 3, or 4 women (since there are only 3 women, 4 is not possible; we correct this below).

However, since there are only 3 women, the possible cases are:

1. Exactly 2 women and 3 men
2. Exactly 3 women and 2 men

We calculate each case separately and sum the results.

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Case 1: Exactly 2 women and 3 men

• Number of ways to choose 2 women from 3: $$C(3,2)=\frac{3!}{2!(3-2)!}=3$$
• Number of ways to choose 3 men from 5: $$C(5,3)=\frac{5!}{3!(5-3)!}=10$$
• Total ways for this case: $$C(3,2)\times C(5,3)=3\times 10=30$$

Case 2: Exactly 3 women and 2 men

• Number of ways to choose 3 women from 3: $$C(3,3)=1$$
• Number of ways to choose 2 men from 5: $$C(5,2)=\frac{5!}{2!(5-2)!}=10$$
• Total ways for this case: $$C(3,3)\times C(5,2)=1\times 10=10$$

Total number of ways:

$$30+10=40$$

Final Answer: The committee can be formed in 40 ways.

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OR: Proof of Combinations Formula and Pascal’s Identity

1. Proof that $C(n,r)=\frac{n!}{(n-r)!r!}$:

The combination formula counts the number of ways to choose $r$ objects from $n$ distinct objects without regard to order.

• The number of permutations of $n$ objects taken $r$ at a time is: $$P(n,r)=\frac{n!}{(n-r)!}$$
• Since combinations ignore order, we divide by the number of ways to arrange $r$ objects, which is $r!$: $$C(n,r)=\frac{P(n,r)}{r!}=\frac{n!}{(n-r)!r!}$$

2. Proof that $C(n,r)+C(n,r-1)=C(n+1,r)$:

This is Pascal’s Identity. We prove it using the combination formula:

$$C(n,r)+C(n,r-1)=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r+1)!(r-1)!}$$

Factor out $\frac{n!}{(n-r+1)!(r-1)!}$:

$$=\frac{n!}{(n-r+1)!(r-1)!}\left[\frac{(n-r+1)}{r}+1\right]$$

Simplify the expression inside the brackets:

$$\frac{(n-r+1)}{r}+1=\frac{(n-r+1)+r}{r}=\frac{n+1}{r}$$

Now, substitute back:

$$=\frac{n!}{(n-r+1)!(r-1)!}\cdot \frac{n+1}{r}=\frac{(n+1)!}{(n+1-r)!r!}=C(n+1,r)$$

Thus, the identity is proven.

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b) Abelian Group and Proof of Inverse Property

Definition of an Abelian Group:

A group $(G,\ast )$ is called abelian (or commutative) if it satisfies the following properties:

1. Closure: For all $a,b\in G$, $a\ast b\in G$.
2. Associativity: For all $a,b,c\in G$, $(a\ast b)\ast c=a\ast (b\ast c)$.
3. Identity Element: There exists an element $e\in G$ such that for all $a\in G$, $a\ast e=e\ast a=a$.
4. Inverse Element: For each $a\in G$, there exists an element $a^{-1}\in G$ such that $a\ast a^{-1}=a^{-1}\ast a=e$.
5. Commutativity: For all $a,b\in G$, $a\ast b=b\ast a$.

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Proof that $(a\ast b{)}^{-1}=a^{-1}\ast b^{-1}$ in an Abelian Group:

Let $(G,\ast )$ be an abelian group, and let $a,b\in G$. We need to show that the inverse of $a\ast b$ is $a^{-1}\ast b^{-1}$.

1. Multiply $a\ast b$ by $a^{-1}\ast b^{-1}$: $$(a\ast b)\ast (a^{-1}\ast b^{-1})$$
2. By associativity, regroup the terms: $$=[a\ast b\ast a^{-1}]\ast b^{-1}$$
3. Since $a\ast a^{-1}=e$, we have: $$=[e\ast b]\ast b^{-1}=b\ast b^{-1}=e$$
4. Similarly, multiply $a^{-1}\ast b^{-1}$ by $a\ast b$: $$(a^{-1}\ast b^{-1})\ast (a\ast b)$$
5. By associativity and commutativity (abelian property): $$=a^{-1}\ast (b^{-1}\ast a)\ast b$$
6. Since $b^{-1}\ast a=a\ast b^{-1}$ (because $G$ is abelian): $$=a^{-1}\ast (a\ast b^{-1})\ast b$$
7. Simplify $a^{-1}\ast a=e$: $$=e\ast b^{-1}\ast b=b^{-1}\ast b=e$$

Since multiplying $a\ast b$ by $a^{-1}\ast b^{-1}$ on both sides yields the identity $e$, we conclude: $$(a\ast b{)}^{-1}=a^{-1}\ast b^{-1}$$