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The no. of words containing 4 alphabets having equal no. of $vowels$ and $consonants$ (repetition of letters allowed) is?

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Since the no. of vowels and consonants are equal,
a word can have 1 vowel and 1 consonant or
2 vowels and 2 consonants.
There are 5 vowels and 21 consonants.
case (1)
No. of words having 1 vowel and 1 consonant
$=^5C_1\times^{21}C_1\times\large\frac{4!}{2!.2!}$$=630$
Case (ii)
No. of words having 2 vowels and 2 consonants
$=^5C_2\times^{21}C_2\times4!$
$\therefore$ The required no. of words
$=630+^5C_2\times^{21}C_2\times 4$
$=630+(10\times 210 \times24)=210(3+240)=210\times 243$
= 51030
answered Sep 3, 2013 by rvidyagovindarajan_1
edited Aug 5, 2014 by sharmaaparna1
 

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