Finding Four-digit numbers

S

stevecowall

New member
Joined
Nov 9, 2011
Messages
34
Question: How many four-digit numbers contain at least one 3 and at least one 7? (we do not consider a number with a leading zero a four-digit number)

My solving:
There are 9000 four-digit positive integers. For those without a 7 and 3, the first digit could be two of the seven numbers 1, 2, 4, 5, 6, 8, or 9, and each of the other digits could be one of the eight numbers 0, 1, 2, 4, 5, 6, 8, or 9. So there are

7 x 8 x 8 x 8 = 3584

I think my solving way is wrong. Can you help me?
 
Last edited:
stevecowall said:
Question: How many four-digit numbers contain at least one 3 and at least one 7? (we do not consider a number with a leading zero a four-digit number)

My solving:
There are 9000 four-digit positive integers. For those without a 7 and 3, the first digit could be two of the seven numbers 1, 2, 4, 5, 6, 8, or 9, and each of the other digits could be one of the eight numbers 0, 1, 2, 4, 5, 6, 8, or 9. So there are

7 x 8 x 8 x 8 = 5416

I think my solving way is wrong. Can you help me?
Click to expand...

I use asterisks, to show multiplication.

7*8*8*8 does not equal 5416.

Also, the expression 7*8*8*8 counts the number of four-digit positive integers that do not contain any 3s or 7s.

You need to count the number of four-digit positive integers that contain both 3s and 7s (at least one of each).

Since two of the four digits must be taken by 3 and 7, how many possibilities are there for each of the two remaining digits?

Remember, the phrase "at least" tells us that there could be more than one 3; there could be more than one 7, too.
 
stevecowall said:
Question: How many four-digit numbers contain at least one 3 and at least one 7? (we do not consider a number with a leading zero a four-digit number)
Click to expand...
You need to know how to use inclusion/exclusion .
There are \(\displaystyle 2\cdot 8\cdot9^3- 7\cdot8^3\) four-digit numbers that contain no 3 or no 7.
Subtract that from 9000 to get the answer.
 
You must log in or register to reply here.
Top