Question

There are five boys A, B, C, D and E. The order of their height is $A < B < C < D < E$. Number of the ways in which they have to be arranged in 4 seats in increasing order of their height such that C & E are never adjacent is
$(a){\text{ 10}} \\\ (b){\text{ 6}} \\\ (c){\text{ 4}} \\\ (d){\text{ 8}} \\\$

Answer 1

Hint – In this problem there are 5 boys and the seats are only 4 thus consider all the possible cases by eliminating one boy in each case. Make sure that the condition of C and E not sitting together satisfies in all these cases formed.

Complete step- by-step solution -
It is given that there are five boys A, B, C, D and E.
The order of their height is $A < B < C < D < E$
Now we have to arrange them in four adjacent seats also in increasing order of their height such that C and E are never adjacent.
Now the possible case are
Case – 1 Eliminate A
So the four boys arranged on four seats in increasing order of their height is
BCDE
Case – 2 Eliminate B
So the four boys arranged on four seats in increasing order of their height is
ACDE
Case – 3 Eliminate C
So the four boys arranged on four seats in increasing order of their height is
ABDE
Case – 4 Eliminate E
So the four boys arranged on four seats in increasing order of their height is
ABCD
As we cannot eliminate D otherwise C and E are adjacent to each other.
So these are the four possible cases in which they have arranged in 4 seats in increasing order of their height.
ABCD, BCDE, ABDE, ACDE

Hence option (C) is correct.

Note – The trick point here was the elimination of the case in which we have to remove D, if this case would have been taken into consideration then the answer would have changed to 5 but that would be wrong. We can’t remove D because C is of lesser height in comparison to D and E is of greater height in comparison to D so if D is removed, the only boy left greater than the height of C would have been E and then we have to place them together but they can’t be placed together, hence this case is eliminated.