Question

How many 7-digit numbers can be formed using the digits 1, 2,0,2,4, 2 and 4?​

Answer 1

Answer:

360

Explanation:

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

Number of 7 digit numbers = 7!3!×2! = 420

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

=6!3!×2! = 60

Hence the required number of 7 digits numbers = 420 - 60 = 360

Answer 2

We've to find no. of 7 - digit numbers made by the 7 digits - 1, 2, 0, 2, 4, 2, 4.

In these 7 digits we see 2 occurs 3 times and 4 occurs 2 times.

Then, no. of such 7 digit numbers expected is,

$\longrightarrow \dfrac{7!}{3!\cdot2!}=420$

But the 7 digits contain one 0, so it can't be at left most digit of our 7 digit numbers.

So we've to deduct no. of possible 7 digit numbers in which 0 occupies left most place, from total no. of numbers.

If 0 occupies left most place, we've 6 digits remaining to be permuted, in which 2 occurs 3 times and 4 occurs 2 times.

Then no. of such numbers is,

$\longrightarrow \dfrac{6!}{3!\cdot2!}=60$

Hence, total no. of possible 7 digit numbers is,

$\longrightarrow 420-60=\bf{360}$