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Question Number 178173 by cortano1 last updated on 13/Oct/22

Answered by Acem last updated on 13/Oct/22

 No.1, NumWrds_(5ℓs) = P_(21) ^( 3) ×P_5 ^( 2) = 159 600 words     No.2, NumWays_(4men) = 2.C_2 ^( 4) = 12 ways

$$\:{No}.\mathrm{1},\:{NumWrds}_{\mathrm{5}\ell{s}} =\:{P}_{\mathrm{21}} ^{\:\mathrm{3}} ×{P}_{\mathrm{5}} ^{\:\mathrm{2}} =\:\mathrm{159}\:\mathrm{600}\:{words} \\ $$$$ \\ $$$$\:{No}.\mathrm{2},\:{NumWays}_{\mathrm{4}{men}} =\:\mathrm{2}.{C}_{\mathrm{2}} ^{\:\mathrm{4}} =\:\mathrm{12}\:{ways} \\ $$

Commented by mr W last updated on 14/Oct/22

No. 1 is wrong sir!  what you got is only the number of   words in a fixed form like CCCVV .  what about CVVCC, VCVCC, etc.?    so correct is: we select at first 3  consonants, there are C_3 ^(21)  ways, and  2 vowels, there are C_2 ^5  ways. with these  5 letters we can form 5! words.  therefore the answer is C_3 ^(21) ×C_2 ^5 ×5!   words.

$${No}.\:\mathrm{1}\:{is}\:{wrong}\:{sir}! \\ $$$${what}\:{you}\:{got}\:{is}\:{only}\:{the}\:{number}\:{of}\: \\ $$$${words}\:{in}\:{a}\:{fixed}\:{form}\:{like}\:{CCCVV}\:. \\ $$$${what}\:{about}\:{CVVCC},\:{VCVCC},\:{etc}.? \\ $$$$ \\ $$$${so}\:{correct}\:{is}:\:{we}\:{select}\:{at}\:{first}\:\mathrm{3} \\ $$$${consonants},\:{there}\:{are}\:{C}_{\mathrm{3}} ^{\mathrm{21}} \:{ways},\:{and} \\ $$$$\mathrm{2}\:{vowels},\:{there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{5}} \:{ways}.\:{with}\:{these} \\ $$$$\mathrm{5}\:{letters}\:{we}\:{can}\:{form}\:\mathrm{5}!\:{words}. \\ $$$${therefore}\:{the}\:{answer}\:{is}\:{C}_{\mathrm{3}} ^{\mathrm{21}} ×{C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{5}!\: \\ $$$${words}. \\ $$

Commented by cortano1 last updated on 14/Oct/22

yes sir

$$\mathrm{yes}\:\mathrm{sir} \\ $$

Commented by Acem last updated on 14/Oct/22

   I think we all were wrong because we forgot the   repeating of letters     the solve must be 21^3 ×5^2 ×5!= 27 783 000

$$ \\ $$$$\:{I}\:{think}\:{we}\:{all}\:{were}\:{wrong}\:{because}\:{we}\:{forgot}\:{the} \\ $$$$\:{repeating}\:{of}\:{letters} \\ $$$$ \\ $$$$\:{the}\:{solve}\:{must}\:{be}\:\mathrm{21}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{2}} ×\mathrm{5}!=\:\mathrm{27}\:\mathrm{783}\:\mathrm{000} \\ $$$$ \\ $$

Commented by mr W last updated on 14/Oct/22

this is wrong!  the question requests that each word   should have different letters.

$${this}\:{is}\:{wrong}! \\ $$$${the}\:{question}\:{requests}\:{that}\:{each}\:{word}\: \\ $$$${should}\:{have}\:{different}\:{letters}. \\ $$

Commented by Acem last updated on 14/Oct/22

Nope, it doesn′t refer to the use different letters

$${Nope},\:{it}\:{doesn}'{t}\:{refer}\:{to}\:{the}\:{use}\:{different}\:{letters} \\ $$

Commented by Acem last updated on 14/Oct/22

Oh , honestly i thought the 3rd line is coming with   the 2nd question

$${Oh}\:,\:{honestly}\:{i}\:{thought}\:{the}\:\mathrm{3}{rd}\:{line}\:{is}\:{coming}\:{with} \\ $$$$\:{the}\:\mathrm{2}{nd}\:{question} \\ $$

Commented by Acem last updated on 14/Oct/22

Please separete between questions with empty   line (:

$${Please}\:{separete}\:{between}\:{questions}\:{with}\:{empty} \\ $$$$\:{line}\:\left(:\right. \\ $$

Commented by mr W last updated on 14/Oct/22

how do you understand “all letters in  each word formed being different”?

$${how}\:{do}\:{you}\:{understand}\:``{all}\:{letters}\:{in} \\ $$$${each}\:{word}\:{formed}\:{being}\:{different}''? \\ $$

Commented by Acem last updated on 14/Oct/22

I haven′t seen it at all before, I′ve just seen it   when you assured about different letters.

$${I}\:{haven}'{t}\:{seen}\:{it}\:{at}\:{all}\:{before},\:{I}'{ve}\:{just}\:{seen}\:{it} \\ $$$$\:{when}\:{you}\:{assured}\:{about}\:{different}\:{letters}. \\ $$

Commented by Tawa11 last updated on 14/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 13/Oct/22

(1)  C_3 ^(21) ×C_2 ^5 ×5!=1596000  (2)  C_2 ^4 ×2=12

$$\left(\mathrm{1}\right) \\ $$$${C}_{\mathrm{3}} ^{\mathrm{21}} ×{C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{5}!=\mathrm{1596000} \\ $$$$\left(\mathrm{2}\right) \\ $$$${C}_{\mathrm{2}} ^{\mathrm{4}} ×\mathrm{2}=\mathrm{12} \\ $$

Commented by Acem last updated on 13/Oct/22

 For the 1st, i think we have to use Permutations

$$\:{For}\:{the}\:\mathrm{1}{st},\:{i}\:{think}\:{we}\:{have}\:{to}\:{use}\:{Permutations} \\ $$

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