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=`...........

Answer

4

Explanation

There are `.^(40)C_(4)` ways of drawing 4 slips `.^(10)C_(2)` ways to choose `a` & `b`. After `a` & `b` have been chosen thre are `.^(4)C_(2)` ways of choosing 2 slips labelled `a` `.^(4)C_(2)` ways of choosingn 2 slips labelled `b`
`:.` Required Probability `=(.^(10)C_(2)xx.^(4)C_(2)xx.^(4)C_(2))/(.^(40)C_(4))=(36.^(10)C_(2))/(.^(40)C_(4))`