- Home
- Standard 11
- Mathematics
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
Similar Questions
normal