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hqi1321
 2 years ago
How many 6 digit positive integers are there, consisting of exactly one 3, one 5, two 4s, and the remaining two digits either 7 or 8?
hqi1321
 2 years ago
How many 6 digit positive integers are there, consisting of exactly one 3, one 5, two 4s, and the remaining two digits either 7 or 8?

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mathmate
 2 years ago
Best ResponseYou've already chosen the best response.1If all the 6 digits are all nonzero and distinct, there are 6! possible integers. If there are two identical digits, we need to divide by 2! Don't know if the 7 and 8's mean A. EITHER two 7's OR two 8's,, or B. EACH digit is 7 or 8. Case A: 1. if both are 7's => 6!/(2!2!) 2. if both are 8's => 6!/(2!2!) Total 2*6!/(2!2!) Case B: 1. same digits (7/7 or 8/8) => 2*6!/(2!2!) 2. different digits (7/8 or 8/7) => 6!/2! Total = 2*6!/(2!2!) + 6!/2! =6!

hqi1321
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for your answer! Sorry, it meant each. Paraphrased a little. I'm confused, why do we divide by 2! if there are 2 identical digits? Why do we divide (why isn't it 6!/(2!2!)
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