Question

Two red counters, three green counters and 4 blue counters are placed in a row in random order. The probability that no two blue counters are adjacent is :-

• A. $\frac{7}{99}$
• B. $\frac{7}{198}$
• C. $\frac{5}{42}$
• D. None of the above

Answer 1

The correct option is C $\frac{5}{42}$

Given,

$2R+3G+4B=9$

Red or Green counters denoted as 0

,

Permissible places for blue as crosses

x 0 x 0 x 0 x 0 x 0 x

far case ${=}^6{C}_4$

total case ${=}^9{C}_4$

$P=\frac{{}^6{C}_4}{{}^9{C}_4}=\frac{15}{126}=\frac{5}{42}$