14.Probability
normal

There are $10$ engineering colleges and five students $A, B, C, D, E$ . Each of these students got offer from all of these $10$ engineering colleges. They randomly choose college independently of each other. Tne probability that all get admission in different colleges can be expressed as $\frac {a}{b}$ where $a$ and $b$ are co-prime numbers then the value of $a + b$ is

A

$814$

B

$731$

C

$1013$

D

$502$

Solution

Total no. of cases $=10^{5}$

All students want different colleges so no. of

cases $ = {\,^{10}}{{\rm{C}}_5} \times 5!$

Probability $ = \,\frac{{^{10}{{\rm{C}}_5} \times 5!}}{{{{10}^5}}} = \frac{{189}}{{625}} = \frac{{\rm{a}}}{{\rm{b}}}$

$a+b=189+625=814$

Standard 11
Mathematics

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