Question ID: #342
The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is:
Solution:
Let $n(A)$ be the number of ways **all boys sit together**.
Treat the 5 boys as a single unit ($B_5$). We have 4 girls + 1 unit of boys = 5 entities.
Arrangements of entities = $5!$.
Internal arrangements of boys = $5!$.
$$n(A) = 5! \times 5! = 120 \times 120 = 14400$$
Let $n(B)$ be the number of ways **no two boys sit together**.
We place the girls first to create gaps:
$$- G_1 – G_2 – G_3 – G_4 -$$
There are 4 girls, arranged in $4!$ ways.
There are 5 gaps created (including ends). We need to place 5 boys in these 5 gaps.
Number of ways to place boys = ${}^5P_5 = 5!$.
$$n(B) = 4! \times 5! = 24 \times 120 = 2880$$
The two cases (all boys together vs no boys together) are mutually exclusive (disjoint).
Total ways = $n(A) + n(B)$
$$= 14400 + 2880$$
$$= 17280$$
Ans. (17280)
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