Question

Shubham has to make a telephone call to his friend Nisheeth, Unfortunately he does not remember the $7$ digit phone number. But he remembers that the first three digits are $635$ or $674$, the number is odd and there is exactly one $9$ in the number. The maximum number of trials that Shubham has to make to be successful is a four digit number of the form $abcd$ then $c$ equal.

Answer 1

There are $2$ ways for arrangement of last $4$ digits

i) $7^{th}$ places i.e last places with $9$ and $4^{th},5^{th},6^{th}$ place with numbers $0$ to $8$

$\therefore$ Permutation for above condition $=9\times 9\times =729$

ii) Not taking $9$ at $7^{th}$ place putting $1,3,5$ or $7$ any one of them at $7^{th}$ place has $4$ ways and their arrangement has $3$ ways and similar in left three place one place will be for $9$ and other $2$ place will be arranged in $=9\times 9=81$

$\therefore$ Permutation for above criteria $=81\times 4\times 3=972$

$\Rightarrow$ Adding both $=729+972=1701$

First $3$ places can be $635$ or $674$ so $2$ ways

$\therefore$ successful or total attempt $=2\times 1701$

$=3402$ ways

$\therefore abcd\Rightarrow 3402$

$x\Rightarrow 0$