Question

If \(m\) things are distributed among \(a\) men and \(b\) women, then the chance that the number of things received by men is odd is:

• A. \(\frac{(b-a)^m - (b+a)^m}{2(b+a)^m}\)
• B. \(\frac{(b+a)^m - (b-a)^m}{2(b+a)^m}\)
• C. \(\frac{(b+a)^m - (b-a)^m}{(b+a)^m}\)
• D. None of these

Hint:

Odd-even separation uses \((x+y)^n \pm (x-y)^n\).

Answer 1

The Correct Option is B

To find the probability that the number of things received by men is odd when \( m \) things are distributed among \( a \) men and \( b \) women, we will use a combinatorial approach to calculate the possible outcomes.

First, let's understand the total possible ways to distribute \( m \) things among \( a + b \) people (i.e., both men and women):

The total ways of distributing \( m \) things among \( a + b \) people can be represented by \((b+a)^m\). This is because each of the \( m \) items can be independently given to any of the \( a + b \) individuals.

\((b+a)^m\) represents the expression for the total number of outcomes.

Now, we need to find the number of ways in which the number of things received by the \( a \) men is odd. To apply the principle of complementary counting and binomial theorem, consider the expression:

\((b-a)^m + (b+a)^m\)

• The expression \((b-a)^m\) gives us the fictitious scenarios where the contributions by men are offset, treating instances where men have identical numbers of items, simplified in fiction.

We derive the probability by subtracting the fictitious equivalences \((b-a)^m\) from the full distribution and then halving non-symmetric cases:

\(\frac{(b+a)^m - (b-a)^m}{2}\)

This expression now represents the favorable scenario where the number of items received by men is odd. To achieve probability, we divide this by the total number of outcomes:

\(\frac{(b+a)^m - (b-a)^m}{2(b+a)^m}\)

This matches option B:

Correct Answer: \(\frac{(b+a)^m - (b-a)^m}{2(b+a)^m}\)