SOLUTION: Q. How many 7-digit numbers are there containing two 3s and three 8s if neither of the other two digits are 3 or 8 (the first digit cannot be zero)?
I tried 7!/5!2!.7!4!3!.7.8 =
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Question 55006This question is from textbook
: Q. How many 7-digit numbers are there containing two 3s and three 8s if neither of the other two digits are 3 or 8 (the first digit cannot be zero)?
I tried 7!/5!2!.7!4!3!.7.8 = 41160
The answer at the back of the book is 12,960. I do not know how that comes about. Please help.
This question is from textbook
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Ignoring the fact that the 1st digit cannot be zero you have
(7C2)(5C3)*8*8 = 13440 7-digit numbers with 2-3's and 3-8's and
any of the other 8 digits in the last two places.
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But some of these numbers DO have zero as the 1st digit; how many??
This many: one way to choose the 1st digit
then (6C2)(4C3)*8= 480 to fill in the next 6 places.
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Final Answer 13440-480 = 12960
Cheers,
Stan H.
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