Question

If $4$ letter words are formed using letters of the word ‘MORADABAD’, then the probability that D comes exactly once in the $4$ letter word is

• A. $\frac{146}{313}$
• B. $\frac{213}{313}$
• C. $\frac{167}{313}$
• D. $\frac{520}{621}$

Answer 1

Answer: (A)

Total cases $=4$ letter word is formed by using either $3$ identical letter and $1$ distinct letter or $2$ alike of one kind and $2$ alike of $2^{nd}$ kind or $2$ alike, $2$ distinct or $4$ distinct
Number of ways for $3$ alike and $1$ distinct $=\_{}^1{C}_1\times \_{}^5{C}_1\times \frac{4!}{3!}=20$
Number of ways for $2$ alike and $2$ alike $=\_{}^2{C}_2\times \frac{4!}{2!2!}=6$
Number of ways for $2$ alike and $2$ distinct $=\_{}^2{C}_1\times \_{}^5{C}_2\times \frac{4!}{2!}=240$
Number of ways for $4$ distinct $=\_{}^6{C}_4\times 4!=360$
The total number of ways $=20+6+240+360=626$
For favorable cases, the following possibilities are there (one D is always included)
Number of ways for $3$ alike $=\_{}^1{C}_1^{}\times \frac{4!}{3!}=4$
Number of ways for $2$ alike $+1$ distinct $=\_{}^1{C}_1^{}\times \_{}^4{C}_1^{}\times \frac{4!}{2!}=48$
Number of ways for $3$ distinct $=\_{}^5{C}_3^{}\times 4!=240$
The total number of ways for favorable cases $=240+48+4=292$
Hence, the required probability $=\frac{292}{626}=\frac{146}{313}$