how many words of $4$ consonants and $3$ vowels can be made from $12$ consonants and $4$ vowels? if all the letters are different?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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my effort for this question is



I am selecting $4$ consonants from $12$ available consonants and $3$ from $4$ available vowels. After selecting $4$ consonants and $3$ vowels now I have $7$ letters in my hand now I am permuting them all with $7!$.
So the total number of words can be made is $12choose44choose37!$.



But the answer is $12choose44choose3$.



I would appreciate if anyone advises on this question.




discrete-mathematics permutations combinations




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  • Weird, at least for me it seems your answer is correct. Maybe the given answer assumes that the consonants need to stay in their own cluster before a cluster of vowels. Odd.
    – Matti P.
    Aug 14 at 7:35






  • 1




    Your answer is correct.
    – N. F. Taussig
    Aug 14 at 7:44














up vote
2
down vote

favorite












my effort for this question is



I am selecting $4$ consonants from $12$ available consonants and $3$ from $4$ available vowels. After selecting $4$ consonants and $3$ vowels now I have $7$ letters in my hand now I am permuting them all with $7!$.
So the total number of words can be made is $12choose44choose37!$.



But the answer is $12choose44choose3$.



I would appreciate if anyone advises on this question.




discrete-mathematics permutations combinations




share|cite|improve this question






















  • Weird, at least for me it seems your answer is correct. Maybe the given answer assumes that the consonants need to stay in their own cluster before a cluster of vowels. Odd.
    – Matti P.
    Aug 14 at 7:35






  • 1




    Your answer is correct.
    – N. F. Taussig
    Aug 14 at 7:44












up vote
2
down vote

favorite









up vote
2
down vote

favorite











my effort for this question is



I am selecting $4$ consonants from $12$ available consonants and $3$ from $4$ available vowels. After selecting $4$ consonants and $3$ vowels now I have $7$ letters in my hand now I am permuting them all with $7!$.
So the total number of words can be made is $12choose44choose37!$.



But the answer is $12choose44choose3$.



I would appreciate if anyone advises on this question.




discrete-mathematics permutations combinations




share|cite|improve this question














my effort for this question is



I am selecting $4$ consonants from $12$ available consonants and $3$ from $4$ available vowels. After selecting $4$ consonants and $3$ vowels now I have $7$ letters in my hand now I am permuting them all with $7!$.
So the total number of words can be made is $12choose44choose37!$.



But the answer is $12choose44choose3$.



I would appreciate if anyone advises on this question.





discrete-mathematics permutations combinations





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 14 at 7:37









apanpapan3

125211




125211










asked Aug 14 at 7:16









Dharmendra Parmar

362




362











  • Weird, at least for me it seems your answer is correct. Maybe the given answer assumes that the consonants need to stay in their own cluster before a cluster of vowels. Odd.
    – Matti P.
    Aug 14 at 7:35






  • 1




    Your answer is correct.
    – N. F. Taussig
    Aug 14 at 7:44
















  • Weird, at least for me it seems your answer is correct. Maybe the given answer assumes that the consonants need to stay in their own cluster before a cluster of vowels. Odd.
    – Matti P.
    Aug 14 at 7:35






  • 1




    Your answer is correct.
    – N. F. Taussig
    Aug 14 at 7:44















Weird, at least for me it seems your answer is correct. Maybe the given answer assumes that the consonants need to stay in their own cluster before a cluster of vowels. Odd.
– Matti P.
Aug 14 at 7:35




Weird, at least for me it seems your answer is correct. Maybe the given answer assumes that the consonants need to stay in their own cluster before a cluster of vowels. Odd.
– Matti P.
Aug 14 at 7:35




1




1




Your answer is correct.
– N. F. Taussig
Aug 14 at 7:44




Your answer is correct.
– N. F. Taussig
Aug 14 at 7:44










1 Answer
1






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up vote
2
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I think you are correct



To take a simpler example, with two consonants from three $R,S,T$ and one vowel from two $A,I$, there are $3C2 cdot 2C1 cdot 3! = 36$ possibilities namely $$RSA, RSI,STA,STI,TRA,TRI,SRA,SRI,TSA,TSI,RTA,RTI,$$ $$RAS, RIS,SAT,SIT,TAR,TIR,SAR,SIR,TAS,TIS,RAT,RIT,$$ $$ARS, IRS,AST,IST,ATR,ITR,ASR,ISR,TSA,ITS,ART,IRT$$



not $3P2 cdot 2P1 = 12$, which would just be the first row in my example of having all the consonants first and then all the vowels






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  • +1 Very convincing. Counting is enough.
    – drhab
    Aug 14 at 9:24










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













I think you are correct



To take a simpler example, with two consonants from three $R,S,T$ and one vowel from two $A,I$, there are $3C2 cdot 2C1 cdot 3! = 36$ possibilities namely $$RSA, RSI,STA,STI,TRA,TRI,SRA,SRI,TSA,TSI,RTA,RTI,$$ $$RAS, RIS,SAT,SIT,TAR,TIR,SAR,SIR,TAS,TIS,RAT,RIT,$$ $$ARS, IRS,AST,IST,ATR,ITR,ASR,ISR,TSA,ITS,ART,IRT$$



not $3P2 cdot 2P1 = 12$, which would just be the first row in my example of having all the consonants first and then all the vowels






share|cite|improve this answer




















  • +1 Very convincing. Counting is enough.
    – drhab
    Aug 14 at 9:24














up vote
2
down vote













I think you are correct



To take a simpler example, with two consonants from three $R,S,T$ and one vowel from two $A,I$, there are $3C2 cdot 2C1 cdot 3! = 36$ possibilities namely $$RSA, RSI,STA,STI,TRA,TRI,SRA,SRI,TSA,TSI,RTA,RTI,$$ $$RAS, RIS,SAT,SIT,TAR,TIR,SAR,SIR,TAS,TIS,RAT,RIT,$$ $$ARS, IRS,AST,IST,ATR,ITR,ASR,ISR,TSA,ITS,ART,IRT$$



not $3P2 cdot 2P1 = 12$, which would just be the first row in my example of having all the consonants first and then all the vowels






share|cite|improve this answer




















  • +1 Very convincing. Counting is enough.
    – drhab
    Aug 14 at 9:24












up vote
2
down vote










up vote
2
down vote









I think you are correct



To take a simpler example, with two consonants from three $R,S,T$ and one vowel from two $A,I$, there are $3C2 cdot 2C1 cdot 3! = 36$ possibilities namely $$RSA, RSI,STA,STI,TRA,TRI,SRA,SRI,TSA,TSI,RTA,RTI,$$ $$RAS, RIS,SAT,SIT,TAR,TIR,SAR,SIR,TAS,TIS,RAT,RIT,$$ $$ARS, IRS,AST,IST,ATR,ITR,ASR,ISR,TSA,ITS,ART,IRT$$



not $3P2 cdot 2P1 = 12$, which would just be the first row in my example of having all the consonants first and then all the vowels






share|cite|improve this answer












I think you are correct



To take a simpler example, with two consonants from three $R,S,T$ and one vowel from two $A,I$, there are $3C2 cdot 2C1 cdot 3! = 36$ possibilities namely $$RSA, RSI,STA,STI,TRA,TRI,SRA,SRI,TSA,TSI,RTA,RTI,$$ $$RAS, RIS,SAT,SIT,TAR,TIR,SAR,SIR,TAS,TIS,RAT,RIT,$$ $$ARS, IRS,AST,IST,ATR,ITR,ASR,ISR,TSA,ITS,ART,IRT$$



not $3P2 cdot 2P1 = 12$, which would just be the first row in my example of having all the consonants first and then all the vowels







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 14 at 7:38









Henry

93.4k471149




93.4k471149











  • +1 Very convincing. Counting is enough.
    – drhab
    Aug 14 at 9:24
















  • +1 Very convincing. Counting is enough.
    – drhab
    Aug 14 at 9:24















+1 Very convincing. Counting is enough.
– drhab
Aug 14 at 9:24




+1 Very convincing. Counting is enough.
– drhab
Aug 14 at 9:24












 

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