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. $\frac{9}{42}$
• B. $\frac{5}{2}$
• C. $\frac{5}{42}$
• D. none

Answer 1

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

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

Red or Green counters denoted as $0$,
Permissible places for blue as crosses
x 0 x 0 x 0 x 0 x 0 x
Restricted placement of blues $={}^6{c}_4$ ways
unrestricted placement of blues $={}^9{c}_4$ ways

Hence,

Required Probability $=\frac{{}^6{c}_4}{{}^9{c}_4}=\frac{5}{42}.$