The first problem is to find the number of ways to arrange the letters in the word "MATHEMATICS". This involves permutations with repetitions. The second problem is to find the number of ways to form a committee consisting of 3 men and 5 women, selected from 5 men and 11 women respectively. This involves combinations.

1. Problem Description

The first problem is to find the number of ways to arrange the letters in the word "MATHEMATICS". This involves permutations with repetitions.

The second problem is to find the number of ways to form a committee consisting of 3 men and 5 women, selected from 5 men and 11 women respectively. This involves combinations.

2. Solution Steps

Problem 1:

The word "MATHEMATICS" has 11 letters. The letters are:

M - 2 times

A - 2 times

T - 2 times

H - 1 time

E - 1 time

I - 1 time

C - 1 time

S - 1 time

The formula for permutations with repetitions is:

$\frac{n!}{n_1!n_2!...n_k!}$

where $n$ is the total number of items, and $n_1, n_2, ..., n_k$ are the counts of each distinct item.

In our case, $n = 11$, $n_1 = 2$ (for M), $n_2 = 2$ (for A), and $n_3 = 2$ (for T).

So the number of arrangements is:

$\frac{11!}{2!2!2!} = \frac{11!}{8} = \frac{39916800}{8} = 4989600$

Problem 2:

We need to choose 3 men from 5 men and 5 women from 11 women. The number of ways to choose $k$ items from a set of $n$ items is given by the combination formula:

$C(n, k) = \frac{n!}{k!(n-k)!}$

The number of ways to choose 3 men from 5 is:

$C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

The number of ways to choose 5 women from 11 is:

$C(11, 5) = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 11 \times 3 \times 2 \times 7 = 462$

To find the total number of ways to form the committee, we multiply the number of ways to choose men and the number of ways to choose women:

$10 \times 462 = 4620$

3. Final Answer

Problem 1: 4989600

Problem 2: 4620