How many 8 digit numbers are there that contain both a 5 and a 6? Number of 8 digit numbers containing both a 5 and a 6= Number of 8 digit numbers - Number of 8 digit numbers lacking both a 5and a 6=90000000−7∗8^7=75319936 Does this seem right?

cochetezgh 2022-09-11 Answered
How many 8 digit numbers are there that contain both a 5 and a 6?
Number of 8 digit numbers containing both a 5 and a 6
= Number of 8 digit numbers - Number of 8 digit numbers lacking both a 5 and a 6
= 90000000 7 8 7
= 75319936
Does this seem right?
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Answers (2)

Raven Mosley
Answered 2022-09-12 Author has 14 answers
A number not containing both 5 and 6, either doesn't contain 5 or doesn't contain 6 (or both). For example, the number 5 doesn't contain both 5 and 6. However, it does contain 5.
You need to use inclusion-exclusion:
- Start with all 9 10 7 numbers with exactly 8 digits.
- Remove all 8 9 7 numbers not containing 5.
- Remove all 8 9 7 numbers not containing 6.
- Numbers containing neither 5 nor 6 were removed twice, so we add them once to compensate: 7 8 7 .
- In total, the result is 9 10 7 2 8 9 7 + 7 8 7 .
Now let's check ourselves by seeing what happens for one-digit numbers and two-digit numbers. For one-digit numbers we get
9 16 + 7 = 0 ,
and indeed no single-digit number can contain both 5 and 6. For two-digit numbers we get
90 144 + 56 = 2 ,
and the two numbers in question are 56 and 65.

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Felix Cohen
Answered 2022-09-13 Author has 4 answers
A different (more complicated) way to solve this is to enumerate on the possible first locations of 5 and 6, calculate explicitly how many numbers there are of this form (easy), and then summing everything up (hard). This is simpler to do if you don't care about "exactly 8 digits" (put otherwise, it's simpler to calculate the number for "up to 8 digits" and then subtract the analogous count for "up to 7 digits").

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