Question

$5$ boys are to be arranged in a row. It two particular boys desire to sit in end places, the number of possible arrangements is
A) $60$
B) $120$
C) $240$
D) $12$

Answer 1

To find the number of possible arrangements while two boys are sitting in a particular seat can be divided into two parts with two boys sitting in both alternative ends and the probability of the other three sitting in the other three mid seats.

Complete step by step solution:
Now, the total number of boys are given as $5$.
Let us take the two boys whose position are fixed as X and Y, if X is sitting in the right end and Y is sitting at the left end then the remaining three places will have the sitting arrangement as $\left( \text{Numbe}{{\text{r}}_{\text{boys in middle}}} \right)!=3!$
Hence, the total number of possibilities is $1\times 3!=1\times 3\times 2\times 1$
$=1\times 3\times 2\times 1$
$=6$
Again, Let us change the two boys positions as Y and X, with X is sitting in the right end and Y is sitting at the left end then the remaining three places will have the sitting arrangement as $\left( \text{Numbe}{{\text{r}}_{\text{boys in middle}}} \right)!=3!$
Hence, the total number of possibilities is $1\times 3!=1\times 3\times 2\times 1$
$=1\times 3\times 2\times 1$
$=6$
Hence, the total number of possibilities with two particular boys sitting at the ends is
$=6+6$
$=12$
$\therefore$ The number of possible arrangements is $12$.

So, correct option is D

Note: Another method to solve this question is by taking variables for each of the names as A, B, C, D, E and taking them into cases with A and E being the two particular boys who sit in the ends.
Case 1: A in left end and E in right end.
Now the rest three can sit as BCD, BDC, CDB, CBD, DBC, DCB making the sitting arrangement of the five boys as ABCDE, ABDCE, ACDBE, ACBDE, ADBCE, and ADCBE. Total of probability is 6.

Case 2: E in left end and A in right end.
Now the rest three can sit as BCD, BDC, CDB, CBD, DBC, DCB making the sitting arrangement of the five boys as EBCDE, EBDCA, ECDBA, EBCDA, EDBCA, and EDCBA. Total of probability is 6.
Therefore, the total probability of sitting is:
$=6+6$
$=12$