Question

Four boys and three girls stand in a queue for an interview, probability that they will in alternate position is

• A. $\frac { 1 } { 34 }$
• B. $\frac { 1 } { 35 }$
• C. $\frac { 1 } { 17 }$
• D. $\frac { 1 } { 68 }$

Answer 1

Answer: (B)

$\frac { 1 } { 35 }$

Answer 2

Four boys can be arranged in $4 !$ ways and three girls can be arranged in 3! ways.

$\therefore$ The favourable cases $= 4 ! \times 3 !$

Hence the required probability $\frac { = 4 ! \times 3 ! } { 7 ! } = \frac { 6 } { 7 \times 6 \times 5 } = \frac { 1 } { 35 }$ .