a) In how many ways a committee of five can be formed out of 4 men and 3 women so that it includes at least one women? [4] b) If a and b are the elements of a group (G, ) , then prove that a x = b…

a) Number of ways to form a committee of five with at least one woman

Given:

• 4 men and 3 women.
• Committee size = 5.
• Must include at least one woman.

Approach: We use the complementary counting method:

1. Calculate the total number of ways to form a committee of 5 from 7 people (4 men + 3 women).
2. Subtract the number of ways to form a committee with no women (i.e., all men).

Step 1: Total number of ways to form a committee of 5

The total number of ways to choose 5 people from 7 is given by the combination formula: $$\text{Total ways}=\left(\genfrac{}{}{0ex}{}{7}{5}\right)=\frac{7!}{5!\cdot (7-5)!}=\frac{7\times 6}{2\times 1}=21$$

Step 2: Number of ways to form a committee with no women (all men)

Since there are 4 men, the number of ways to choose 5 men is: $$\left(\genfrac{}{}{0ex}{}{4}{5}\right)=0$$ (This is zero because we cannot choose 5 men from only 4.)

However, if the committee size were ≤ 4, we would calculate it. But since the committee must have at least one woman, we must consider cases where the committee has 1, 2, or 3 women (since there are only 3 women).

Alternative Approach: Direct Counting

We calculate the number of ways to form a committee with:

1. Exactly 1 woman and 4 men
2. Exactly 2 women and 3 men
3. Exactly 3 women and 2 men

Case 1: 1 woman and 4 men

• Choose 1 woman from 3: $\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3$
• Choose 4 men from 4: $\left(\genfrac{}{}{0ex}{}{4}{4}\right)=1$
• Total ways: $3\times 1=3$

Case 2: 2 women and 3 men

• Choose 2 women from 3: $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$
• Choose 3 men from 4: $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$
• Total ways: $3\times 4=12$

Case 3: 3 women and 2 men

• Choose 3 women from 3: $\left(\genfrac{}{}{0ex}{}{3}{3}\right)=1$
• Choose 2 men from 4: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$
• Total ways: $1\times 6=6$

Total number of valid committees: $$3+12+6=21$$

But wait! The total number of ways to form a committee of 5 from 7 is 21, and the number of ways to form a committee with no women is 0 (since we cannot choose 5 men from 4). Thus, all 21 committees must include at least one woman.

However, this seems counterintuitive because we cannot form a committee of 5 men from 4. Therefore, every possible committee of 5 must include at least one woman.

But let’s verify:

• Maximum men in a committee of 5 = 4 (since there are only 4 men).
• Thus, the committee must have at least 1 woman.

Final Answer: The number of ways to form a committee of 5 with at least one woman is 21.

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b) Proving uniqueness of solutions in a group

Given:

• $(G,\ast )$ is a group.
• $a,b\in G$.
• We need to prove that the equations $a\ast x=b$ and $x\ast a=b$ have unique solutions in $G$.

Proof for $a\ast x=b$:

1. Existence of a solution:

   • Since $(G,\ast )$ is a group, every element has an inverse.
   • Multiply both sides of $a\ast x=b$ on the left by $a^{-1}$: $$a^{-1}\ast (a\ast x)=a^{-1}\ast b$$
   • By associativity: $$(a^{-1}\ast a)\ast x=a^{-1}\ast b$$
   • Since $a^{-1}\ast a=e$ (identity element): $$e\ast x=a^{-1}\ast b⟹x=a^{-1}\ast b$$
   • Thus, $x=a^{-1}\ast b$ is a solution.

2. Uniqueness of the solution:

   • Suppose there exists another solution $x^{\prime }$ such that $a\ast x^{\prime }=b$.
   • Then: $$a\ast x^{\prime }=b=a\ast x$$
   • Multiply both sides on the left by $a^{-1}$: $$a^{-1}\ast (a\ast x^{\prime })=a^{-1}\ast (a\ast x)$$
   • Simplifying: $$x^{\prime }=x$$
   • Hence, the solution is unique.

Proof for $x\ast a=b$:

1. Existence of a solution:

   • Multiply both sides of $x\ast a=b$ on the right by $a^{-1}$: $$(x\ast a)\ast a^{-1}=b\ast a^{-1}$$
   • By associativity: $$x\ast (a\ast a^{-1})=b\ast a^{-1}$$
   • Since $a\ast a^{-1}=e$: $$x\ast e=b\ast a^{-1}⟹x=b\ast a^{-1}$$
   • Thus, $x=b\ast a^{-1}$ is a solution.

2. Uniqueness of the solution:

   • Suppose there exists another solution $x^{\prime }$ such that $x^{\prime }\ast a=b$.
   • Then: $$x^{\prime }\ast a=b=x\ast a$$
   • Multiply both sides on the right by $a^{-1}$: $$(x^{\prime }\ast a)\ast a^{-1}=(x\ast a)\ast a^{-1}$$
   • Simplifying: $$x^{\prime }=x$$
   • Hence, the solution is unique.

Conclusion: Both equations $a\ast x=b$ and $x\ast a=b$ have unique solutions in $G$, given by: $$x=a^{-1}\ast b\text{and}x=b\ast a^{-1}$$ respectively.

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OR

i) Proving $(a\ast b{)}^{-1}=b^{-1}\ast a^{-1}$

Given:

• $(G,\ast )$ is a group.
• $a,b\in G$.

Proof: We need to show that $b^{-1}\ast a^{-1}$ is the inverse of $a\ast b$.

1. Multiply $(a\ast b)$ by $(b^{-1}\ast a^{-1})$: $$(a\ast b)\ast (b^{-1}\ast a^{-1})=a\ast (b\ast b^{-1})\ast a^{-1}$$
2. Since $b\ast b^{-1}=e$ (identity): $$a\ast e\ast a^{-1}=a\ast a^{-1}=e$$
3. Similarly, multiply $(b^{-1}\ast a^{-1})$ by $(a\ast b)$: $$(b^{-1}\ast a^{-1})\ast (a\ast b)=b^{-1}\ast (a^{-1}\ast a)\ast b$$
4. Since $a^{-1}\ast a=e$: $$b^{-1}\ast e\ast b=b^{-1}\ast b=e$$
5. Thus, $b^{-1}\ast a^{-1}$ satisfies the definition of the inverse of $a\ast b$.

Conclusion: $$(a\ast b{)}^{-1}=b^{-1}\ast a^{-1}$$

ii) Proving $(a^{-1}{)}^{-1}=a$

Given:

• $a\in G$, where $(G,\ast )$ is a group.

Proof:

1. By definition of inverse, $a^{-1}$ satisfies: $$a\ast a^{-1}=e\text{and}{a}^{-1}\ast a=e$$
2. Now, consider $(a^{-1}{)}^{-1}$. By definition, it must satisfy: $$a^{-1}\ast (a^{-1}{)}^{-1}=e\text{and}(a^{-1}{)}^{-1}\ast a^{-1}=e$$
3. But we already know that $a$ satisfies: $$a^{-1}\ast a=e\text{and}a\ast a^{-1}=e$$
4. Therefore, $(a^{-1}{)}^{-1}$ must be equal to $a$ to satisfy the inverse property.

Conclusion: $$(a^{-1}{)}^{-1}=a$$