Five friends $F_{1}, F_{2}, F_{3}, F_{4}, F_{5}$ book five seats $C_{1}, C_{2}, C_{3}, C_{4}, C_{5}$ respectively of movie KABIL independently (i,e. $F_{1}$ books $C_{1}, F_{2}$ books $C_{2}$ and so on). In how many different ways they can sit on these seats if $F_{1}$ and $F_{2}$ wants to sit adjacent to each other.

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Solution

Given
$F_{1}, F_{2}, F_{3}, F_{4}, F_{5}$ booked five seats $C_{1}, C_{2}, C_{3}, C_{4}, C_{5}$
Now All friends are independent to seat on any seat out 5 seats So
Number of ways are $5!$
But $F_{1}$ and $F_{2}$ wants to seat adjacent seats
Hence No of ways are $^{5} C_{2}$
Hence Total number of ways $=^{5} C_{2} \times 3!$
$w a y s = \frac{5!}{2! 3!} \times 3! = 5 \times 4 \times 3 = 60$

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Five friends F1,F2,F3,F4,F5 book five seats C1,C2,C3,C4,C5 respectively of movie KABIL independently (i,e. F1 books C1,F2 books C2 and so on). In how many different ways they can sit on these seats if F1 and F2 wants to sit adjacent to each other.

Given F1,F2,F3,F4,F5 booked five seats C1,C2,C3,C4,C5Now All friends are independent to seat on any seat out 5 seats So Number of ways are 5!But F1 and F2 wants to seat adjacent seatsHence No of ways are 5C2Hence Total number of ways =5C2×3!ways=5!2!3!×3!=5×4×3=60