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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-> 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
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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