2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Your answer is correct. $\endgroup$ – N. F. Taussig Aug 14 '18 at 7:44
2
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.