A) \[\frac{4}{5}\]
B) 1/5
C) 1/10
D) 9/10
Correct Answer: A
Solution :
[a] Total no. of arrangements of the letters of the word UNIVERSITY is \[\frac{10!}{2!}\]. No. of arrangements when both I?s are together \[=9!\] So. The no. of ways in which 2 I?s do not together \[=\frac{10!}{2!}-9!\] \[\therefore \] Required probability \[=\frac{\frac{10!}{2!}-9!}{\frac{10!}{2!}}=\frac{10!-9!2!}{10!}\] \[=\frac{10\times 9!-9!2!}{10!}=\frac{9![10-2]}{10\times 9!}=\frac{8}{10}=\frac{4}{5}\]You need to login to perform this action.
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