40 slips are place in a box, each bearing a number 1 to 10 with each number entered on 4 slips. 4 slips are drawn from the box at random without replacement. The probability that two of the slips bear a number `'a'` & other two bear a number `'b' (!= a)` is `k/9 xx (.^(10)C_(2))/(.^(40)C_(4))` where `k=`...........

Explanation

To solve the problem, we need to calculate the probability that when drawing 4 slips from a box containing 40 slips (4 slips each for numbers 1 to 10), exactly 2 slips bear the number 'a' and the other 2 slips bear the number 'b' (where 'b' is not equal to 'a').

### Step-by-Step Solution:

1. **Identify the Total Number of Slips**:
There are 40 slips in total, with each number from 1 to 10 represented on 4 slips.

2. **Choose Numbers 'a' and 'b'**:
We need to select 2 different numbers from the 10 available numbers. The number of ways to choose 2 numbers from 10 is given by the combination formula \( \binom{10}{2} \).

\[ \text{Ways to choose } a \text{ and } b = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]

3. **Choose 2 Slips for Number 'a'**:
Since there are 4 slips for each number, the number of ways to choose 2 slips from the 4 slips of number 'a' is \( \binom{4}{2} \).

\[ \text{Ways to choose 2 slips for } a = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \]

4. **Choose 2 Slips for Number 'b'**:
Similarly, the number of ways to choose 2 slips from the 4 slips of number 'b' is also \( \binom{4}{2} \).

\[ \text{Ways to choose 2 slips for } b = \binom{4}{2} = 6 \]

5. **Calculate the Total Favorable Outcomes**:
The total number of favorable outcomes for drawing 2 slips of 'a' and 2 slips of 'b' is the product of the ways to choose 'a' and 'b' and the ways to choose the slips.

\[ \text{Total favorable outcomes} = \binom{10}{2} \times \binom{4}{2} \times \binom{4}{2} = 45 \times 6 \times 6 = 1620 \]

6. **Calculate the Total Possible Outcomes**:
The total number of ways to choose any 4 slips from the 40 slips is given by \( \binom{40}{4} \).

\[ \text{Total possible outcomes} = \binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91,390 \]

7. **Calculate the Probability**:
The probability \( P \) that two slips bear the number 'a' and two slips bear the number 'b' is given by the ratio of favorable outcomes to total outcomes.

\[ P = \frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{1620}{91,390} \]

8. **Express the Probability in the Given Form**:
We need to express this probability in the form \( \frac{k}{9} \times \frac{\binom{10}{2}}{\binom{40}{4}} \).

First, calculate \( \frac{\binom{10}{2}}{\binom{40}{4}} \):

\[ \frac{45}{91,390} \]

Now, we can set up the equation:

\[ \frac{1620}{91,390} = \frac{k}{9} \times \frac{45}{91,390} \]

By cross-multiplying, we get:

\[ 1620 = \frac{k \times 45}{9} \]

Simplifying gives:

\[ k \times 5 = 1620 \implies k = \frac{1620}{5} = 324 \]

### Final Answer:
Thus, the value of \( k \) is \( 324 \).