Math, asked by bangarichowdary, 7 months ago

how many numbers can be formed greater than a millon with one 0 and two 3 's and two 7's and two 1's ​

Answers

Answered by meghanasolanki1977
1

Any number of greater than a million will contain all the seven digits.

Now, we have to arrange these seven digits, out of which 2 occurs twice, 3 occurs twice and the rest are distinct.

The number of such arrangements =

2!×3!

7!

=

2!×3!

7×6×5×4×3!

=7×6×5×2=420

These arrangements also include those numbers which contain 0 at the million’s place.

Keeping 0 fixed at the millionth place, we have 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.

These arrangements also include those numbers which contain 0 at the million’s place.

Keeping 0 fixed at the millionth place, we

=6×5×2=60 ways

Hence, the number of required numbers =420–

Similar questions