How many different 6-digit numbers can be made by using each of the following six digits: 2,5,5,9,9,9 exactly once?

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I am able to solve if the digits are different by using the solution 6x5x4x3x2x1=720 but with the repeated numbers, I am not sure how to apply this method.






combinatorics permutations







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  • 2




    How many ways are there to place the $9's$? After they are placed, how many ways are there to place the $5's$?
    – lulu
    Jun 19 '17 at 9:48














up vote
0
down vote

favorite












I am able to solve if the digits are different by using the solution 6x5x4x3x2x1=720 but with the repeated numbers, I am not sure how to apply this method.






combinatorics permutations







share|cite|improve this question



















  • 2




    How many ways are there to place the $9's$? After they are placed, how many ways are there to place the $5's$?
    – lulu
    Jun 19 '17 at 9:48












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am able to solve if the digits are different by using the solution 6x5x4x3x2x1=720 but with the repeated numbers, I am not sure how to apply this method.






combinatorics permutations







share|cite|improve this question















I am able to solve if the digits are different by using the solution 6x5x4x3x2x1=720 but with the repeated numbers, I am not sure how to apply this method.






combinatorics permutations




combinatorics permutations






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share|cite|improve this question













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edited Sep 10 at 8:28









N. F. Taussig

39.9k93253




39.9k93253










asked Jun 19 '17 at 9:44









Jacqueline Kuek

11




11







  • 2




    How many ways are there to place the $9's$? After they are placed, how many ways are there to place the $5's$?
    – lulu
    Jun 19 '17 at 9:48












  • 2




    How many ways are there to place the $9's$? After they are placed, how many ways are there to place the $5's$?
    – lulu
    Jun 19 '17 at 9:48







2




2




How many ways are there to place the $9's$? After they are placed, how many ways are there to place the $5's$?
– lulu
Jun 19 '17 at 9:48




How many ways are there to place the $9's$? After they are placed, how many ways are there to place the $5's$?
– lulu
Jun 19 '17 at 9:48










1 Answer
1






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1
down vote













You need to divide by $3!$ for the nine's and by $2!$ for the fives, giving $60$.






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  • shouldn't it be 120 then?
    – Jacqueline Kuek
    Jun 19 '17 at 9:51










  • @JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
    – callculus
    Jun 19 '17 at 9:56










  • Ok i think i understand now
    – Jacqueline Kuek
    Jun 19 '17 at 9:57










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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote













You need to divide by $3!$ for the nine's and by $2!$ for the fives, giving $60$.






share|cite|improve this answer




















  • shouldn't it be 120 then?
    – Jacqueline Kuek
    Jun 19 '17 at 9:51










  • @JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
    – callculus
    Jun 19 '17 at 9:56










  • Ok i think i understand now
    – Jacqueline Kuek
    Jun 19 '17 at 9:57














up vote
1
down vote













You need to divide by $3!$ for the nine's and by $2!$ for the fives, giving $60$.






share|cite|improve this answer




















  • shouldn't it be 120 then?
    – Jacqueline Kuek
    Jun 19 '17 at 9:51










  • @JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
    – callculus
    Jun 19 '17 at 9:56










  • Ok i think i understand now
    – Jacqueline Kuek
    Jun 19 '17 at 9:57












up vote
1
down vote










up vote
1
down vote









You need to divide by $3!$ for the nine's and by $2!$ for the fives, giving $60$.






share|cite|improve this answer












You need to divide by $3!$ for the nine's and by $2!$ for the fives, giving $60$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 19 '17 at 9:48









JonMark Perry

11.2k92238




11.2k92238











  • shouldn't it be 120 then?
    – Jacqueline Kuek
    Jun 19 '17 at 9:51










  • @JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
    – callculus
    Jun 19 '17 at 9:56










  • Ok i think i understand now
    – Jacqueline Kuek
    Jun 19 '17 at 9:57
















  • shouldn't it be 120 then?
    – Jacqueline Kuek
    Jun 19 '17 at 9:51










  • @JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
    – callculus
    Jun 19 '17 at 9:56










  • Ok i think i understand now
    – Jacqueline Kuek
    Jun 19 '17 at 9:57















shouldn't it be 120 then?
– Jacqueline Kuek
Jun 19 '17 at 9:51




shouldn't it be 120 then?
– Jacqueline Kuek
Jun 19 '17 at 9:51












@JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
– callculus
Jun 19 '17 at 9:56




@JacquelineKuek $frac6!2!cdot 3!=frac7202cdot 6=60$
– callculus
Jun 19 '17 at 9:56












Ok i think i understand now
– Jacqueline Kuek
Jun 19 '17 at 9:57




Ok i think i understand now
– Jacqueline Kuek
Jun 19 '17 at 9:57

















 

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