If the digits 7 and 4 are used at least once?
i think its P(5,2)*5^3 +P(5,4)*5 is that correct?
+118629 How many 5 digit numbers can be made from the digits 1,3,4,6,7,8,9 if the 4 and 6 must be used at least once.
that is
(1,3,7,8,9) and (4 ,6)
I am going to simplify the question and say that no digit can be used twice and the 4 and 6 must be included.
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So that would be 4 and 6 and 3 other digits
5C3 * 5! = 10*120 = 1200 ways
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Now I will assume you want to allow multiples of any digit. Is that correct? That is harder question.
No 6es= 6^5 ways
No 4s = 6^5 ways
No 4sw and no 6es = 5^5 ways (I will add this back because these will have been taken out twice.)
Any combination = 7^5 ways
Any combination so long as there is at least one 4 and one 6 = 7^5-6^5-6^5+5^5
7^5-6^5-6^5+5^5 = 4380
That is what I think. :))
It is different from your answer.
