Question

A number of four different digits is formed with the help of the digits $1,2,3,4,5,6,7$ in all possible ways. Let
$\left(i\right)$ k be the number of such numbers can be formed.
$\left(ii\right)$ m be number of such numbers which are even .
$\left(iii\right)$ n be number of such numbers which are exactly divisible by $4$.
$\left(iv\right)$ p be number of such numbers which are are exactly divisible by $25$.
Find $\frac{k+m-n}{p}$ ?

Answer 1

Here total digits $=7$ and no two of these are alike.
$\left(i\right)$ Required number of ways = Taking $4$ out of $7$
$={}^7{C}_4$
$=7\times 6\times 5\times 4$
$=840$
$\left(ii\right)$ For even number, last digit must be $2$ or $4$ or $6$. Now the remaining three first places are to be filled from the remaining $6$ digits and this can be done in ${}^6{P}_3=\mathrm{6.5.4}=120$ ways
And the last digit can be filled in $3$ ways
$\therefore$ By the principle of multiplication, the required number of ways
$=120\times 3=360$
$\left(iii\right)$ For the number exactly divisible by $4$, then last two digits must be divisible by $4$, the last two digits are viz. $12,16,24,32,36,52,56,64,72,76$. Total $10$ ways.
Now the remaining two first places on the left of $4$ digit numbers are to be filled from the remaining $5$ digits and this can be done in ${}^5{P}_2=20$ ways.
Hence, by the principle of multiplication, the required number of ways
$=20\times 10$
$=200$
$\left(iv\right)$ For the number exactly divisible by $25$, then last $2$ digits must be divisible by $25$, the last two digits are viz. $25,75$. Total 2 ways.
Now the remaining two first places on the left of the $4$ digit number are to be filled from the remaining $5$ digits and this can be done in ${}^5{P}_2=20$ ways.
Hence, by the principle of multiplication, the required number of ways
$=20\times 2=40$
$\Rightarrow \frac{k+m-n}{p}=25$