Mathematics - Probability Question with Solution | TestHub

Number of girls $=3$, number of boys $=7$. Since there is no restriction on boys, therefore first of all arrange the 7 boys in ${}^7{P}_7=7$ ! ways.

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If the girls are arranged at the places (including the two ends) indicated by crosses, no two of three girls will be consecutive.

Now there are 8 places for 3 girls

3 girls can be arranged in ${}^8{P}_3$ ways Required number

$$={}^8{P}_{3}7!=\frac{8!}{5!}\times 7!=336\times 5040=3\times (564480)$$

Required number