First slide

Combinations

Question

$Six people A,B,C,D,E,F are going to sit in a row on a bench. Let A and B are to be adjacent. C does not want to$ $sit adjacent to D . E and F can sit anywhere. Number of ways in which these six people$ can be seated, is:

Moderate

A

200
B

144
C

120
D

56

Solution

$C o n s i d e r A B a s o n e u n i t . C and D are to be separated; E and F any where$
$A B and E, F can be seated in a row in 3! x 2! ways$
$No. of gaps between AB, E, F is 4 □ A B □ E □ F □$
$In those 4 gaps C, D can be seated in ({}^{4}C_{2} x 2!) ways$
$Total number of ways = (3! x 2!) x ({}^{4}{C_{2}} x 2!) = 144$

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