Question

How many different linear arrangements are there of the letters A, B, C, D, E, F for which

(a) A and B are next to each other?

(b) A is before B?

(c) A is before B and B is before C?

(d) A is before B and C is before D?

(e) A and B are next to each other and C and D are also next to each other?

(f) E is not last in line?

Short answer

(a) The no. of linear arrangements of the letters A, B, C, D, E, F such that A and B are next to each other is$240.$

(b) The no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B is$360$.

(c) The no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B and B is before C is$120.$

(d) The no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B and C is before D is$180$.

(e) The no. of linear arrangements of the letters A, B, C, D, E, F such that A and B are next to each other and C and D are also next to each other is$96$.

(f) The no. of linear arrangements of the letters A, B, C, D, E, F such that E is not last in line is$600$.

Step-by-step solution

Part (a) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A and B are next to each other.

Suppose A and B are together and form a group, then no. of groups are $5$.

These groups can be arranged in $5!\text{ways}=5\times 4\times 3\times 2\times 1=120\text{ways}$

A and B can be arranged in $2!\text{ways}=2\times 1=2\text{ways}$.

Therefore, the possible no. of arrangements are$=120\times 2=240$.

Part (b) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B.

Total no. of arrangements with six letters $6!=6\times 5\times 4\times 3\times 2\times 1=720$

The no. of arrangements in which A is before B $=\frac{720}{2}$

Therefore, the possible no. of arrangements are $=360$.

Part (c) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B and B is before C.

Total no. of arrangements with six letters $6!=6\times 5\times 4\times 3\times 2\times 1=720$

A, B and C can be arranged in $3!\text{ways}=3\times 2\times 1=6\text{ways}$

Out of these 6 arrangements, there is only one arrangement in which A is before B and B is before C.

So, the possible no. of ways = $\left(\begin{array}{c}6\\ 1\end{array}\right)=\frac{6!}{1!5!}=\frac{6\times 5!}{1\times 5!}=6$

Therefore, the possible no. of arrangements are$\frac{720}{6}=120$.

Part (d) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B and C is before D.

The no. of arrangements in which A is before B $=\frac{6!}{2}=360$

Out of these $360$arrangements, half will be with C before D and half will be with D before C.

Therefore, the possible no. of arrangements are$\frac{360}{2}=180$.

Part (e) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A and B are next to each other and C and D are also next to each other.

If A and B form one group and C and D from another group then there are $4$groups and these groups can be arranged in $4!\text{ways}=4\times 3\times 2\times 1=24\text{ways}$

A and B can be arranged in $2!\text{ways}=2\times 1=2\text{ways}$

Similarly, C and D can be arranged in $2!\text{ways}=2\times 1=2\text{ways}$

Therefore, the possible no. of arrangements are$24\times 2\times 2=96$.

Part (f) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that E is not last in line.

The no. of arrangements in which the position of E is fixed at last $=5!=5\times 4\times 3\times 2\times 1=120$

The no. of arrangements in which the position of E is not fixedrole="math" localid="1648405653455" $=6!=6\times 5\times 4\times 3\times 2\times 1=720$

Therefore, the possible no. of arrangements are $720-120=600$.