Counting Combinations & Permutations with Repetition

AI Thread Summary
To determine how many 6-digit numbers greater than 800,000 can be formed from the digits 1, 1, 5, 5, 5, and 8, the first digit must be 8. The remaining digits consist of two 1's and three 5's. The number of distinct arrangements of these digits is calculated using the formula (5!)/(2! * 3!), resulting in 10 different combinations. Clarification on the use of brackets in the formula was also discussed. This approach effectively addresses the problem of counting combinations and permutations with repetition.
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I had mono while this unit was being taught so I am havin quite a lot of trouble figurin this homework out. Like this question:

How many 6 digit numbers greater than 800 000 can be made from the digits 1, 1, 5, 5, 5, 8?

I have absolutly no idea so any help would be appriciated! Thanks!
 
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Format said:
I had mono while this unit was being taught so I am havin quite a lot of trouble figurin this homework out. Like this question:

How many 6 digit numbers greater than 800 000 can be made from the digits 1, 1, 5, 5, 5, 8?

I have absolutly no idea so any help would be appriciated! Thanks!
Because numbers must be greater than (800,000), the first digit must be 8. The number of different arrangements of the remaining 5 digits, consisting of 2-(1)'s and 3-(5)'s is given by:
{Number Arrangements} = (5!)/{(2!)*(3!)} = (10)


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ah ok i was tryin something like that, but i didnt realize brackets were necessary. Thankyou!
 
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