Question

How many 10 digit decimal numbers contain:

a-) exactly three 7’s ?

b-) at most two 7’s ?

c-) at least two 7’s ?

please explain when solving problems. thanks

Answer 1

A 10 digit decimal number must have a digit 1-9 in the first position and all other places can be occupied by any of the 10 digits(0-9).

A.

Exactly three 7's:

We choose three of the 3 of the 10 digits are 7s = 10C3

= 120

Then the other 7 digits can be filled with anything except 7 so 9 choices = 9⁷

Total ways = 120*9⁷

= 573956280

But the above combination will also include the digits which have 0 as the initial digit and other remaining 9 digits will have 3 7's. So,

Total of these no. = 9C3*9⁶

= 84*531441

= 44641044

Thus total numbers with exactly three 7's = 10C3*9⁷-9C3*9⁶ .......1

= 573956280 - 44641044

= 529315236

B.

Almost two 7's = zero 7's + one 7's + two 7's

zero 7's = 8*9⁹ = 3099363912

Using 1:

one 7's = 10C1*9⁹ - 9C1*9⁸ = 3486784401

two 7's = 10C2*9⁸ - 9C2*9⁷ = 1764915561

Total = 8351063874

C.

At least two 7's = Total - zero 7's - one 7's

= 9*10⁹ - 3099363912 - 3486784401

= 2413851687

Please upvote if you have liked my answer, would be of great help. Thank you.