The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7…

Question

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is:

Accepted Answer

We have to form a $7$-digit number using the digits $1,2,3,5,7$ such that each digit is used at least once.

Since there are $5$ distinct digits and the number has $7$ places, after using each digit once, there are $2$ extra places.

So, the repetition can happen in the following ways:

Case 1: One digit is repeated $3$ times
Then the distribution of digits is:

$3,1,1,1,1$

Choose the digit which appears $3$ times:

${}^{5}{C}_{1}=5$

Now arrange the $7$ digits:

$\frac{7!}{3!}$

So, total numbers in this case:

$5	imes \frac{7!}{3!}=5	imes \frac{5040}{6}=5	imes 840=4200$

Case 2: Two digits are repeated $2$ times each
Then the distribution is:

$2,2,1,1,1$

Choose the two digits which are repeated:

${}^{5}{C}_{2}=10$

Now arrange the $7$ digits:

$\frac{7!}{2!2!}$

So, total numbers in this case:

$10	imes \frac{7!}{2!2!}=10	imes \frac{5040}{4}=10	imes 1260=12600$

Total number of such $7$-digit numbers
$4200+12600=16800$

Hence, the required number is

$16800$

So, the correct option is Option C.

Solutions · 1