An urn contains marbles of four colours : red, white, blue and green. When four marbles are drawn without replacement, the following events are equally likely :
(1) the selection of four red marbles
the selection of one white and three red marbles
(3) the selection of one white, one blue and two red marbles
(4) the selection of one marble of each colour
The smallest total number of satisfying the given condition is

Options

A. 19
B. 21
C. 46
D. 69

Answer

B

Explanation

Let Red Balls = x
White Balls = y
Blue Balls = z
Green Balls = w

`(.^(x)C_(4))/(x+y+z+wC_(4))=(.^(x)C_(3)*.^(y)C_(1))/(x+y+z+wC_(4))=(.^(x)C_(2).^(y)C_(1).^(z)C_(1))=(.^(x)C_(1)xx.^(y)C_(1).^(z)C_(1)xx.^(w)C_(1))/(x+y+z+wC_(4))=(x+y+z+wC_(4))`
`.^(x)C_(4)=.^(x)C_(3).^(y)C_(1)" "x-3=4y" "x=4y+3`
`(.^(x)C_(3).^(y)C_(1)=.^(x)C_(2).^(y)C_(1).^(z)C_(1)" "x-2=2z" "x=3z+2`
`.^(x)C_(2).^(y)C_(1)=.^(x)C_(1).^(y)C_(1).^(z)C_(1).^(w)C_(1)" "x-1=2w" "x=2w+1`
Cleary for y=1 not possible
at y=2 x=11
z=3 x=11
w=5 x=11
so, minimum number of Ball =11+2+3+5+21