How many 5-digit numbers are there in which three of the digits are 7, or two of the digits are 2?

Answer 1

We assume that "3 of the digits are 7" means "exactly 2 of the digits are 7" and "2 of the digits are 2" means "exactly 2 of the digits are 2." We also assume that the 5-digit numbers run from 00000 to 99999 ..

Notation:

C(a,b) will mean "a choose b", the number of b-element subsets of a set with a elements C(a,b) is only nonzero if a and b are integers with 0 <= b <= a.

For any finite set X, let n(X) be the number of elements of X.

For two sets X and Y, X U Y is the union of X and Y, and X & Y is the intersection of X and Y.

Let S be the set of 5-digit numbers for which three of the digits are 7 and let T be the set of 5-digit numbers for which two of the digits are 2. Then we want to compute n(S U T), which is n(S) + n(T) - n(S & T).

Now n(S) is:

the number of ways to choose three digit positions from 1 to 5 to place three 7's

times

the number of ways to choose two (possibly equal) non-7' digits to place in the other two positions,

which is

C(5,3) = 10

times

9 x 9 .

So n(S) = 810

And n(T) is:

the number of ways to choose 2 digit positions from 1 to 5 to place two 2's

times

the number of ways to choose three non-two digits (repeating digits allowed) to place into the other three positions.

which is

C(5,2)

times

9 x 9 x 9 ,

So n(T) = 10 x 9 x 9 x 9 = 7,290.

And n(S & T) is the number of ways of placing the two 2 digits. The other digits must be 7's. That's C(5,2)=10.

So C(S U T) = 810 + 7,290 - 10 = 8,090 is the number of 5-digit numbers with three 7's or two 2's.