Counting Combinations & Permutations with Repetition

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SUMMARY

The discussion focuses on calculating the number of 6-digit numbers greater than 800,000 that can be formed using the digits 1, 1, 5, 5, 5, and 8. The first digit must be 8 to satisfy the condition of being greater than 800,000. The remaining digits consist of two 1's and three 5's, leading to the formula for arrangements: (5!)/(2! * 3!) = 10. This calculation confirms that there are exactly 10 valid combinations.

PREREQUISITES
  • Understanding of permutations and combinations
  • Familiarity with factorial notation
  • Basic knowledge of number properties
  • Ability to apply combinatorial formulas
NEXT STEPS
  • Study the principles of permutations with repetition
  • Learn about factorial calculations and their applications
  • Explore combinatorial problems involving constraints
  • Practice with similar problems using different digit sets
USEFUL FOR

Students learning combinatorics, educators teaching mathematical concepts, and anyone interested in solving permutation problems involving constraints.

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

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