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 ${12\choose4}{4\choose3}7!$.

But the answer is ${12\choose4}{4\choose3}$.

I would appreciate if anyone advises on this question.

  • $\begingroup$ 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. $\endgroup$ – Matti P. Aug 14 '18 at 7:35
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    $\begingroup$ Your answer is correct. $\endgroup$ – N. F. Taussig Aug 14 '18 at 7:44

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

  • $\begingroup$ +1 Very convincing. Counting is enough. $\endgroup$ – drhab Aug 14 '18 at 9:24

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