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Permutations and Combinations
The total no. of words that can be formed using the alphabets $a,b,c,d,e,f$ taking 3 at a time so that each word has atleast one vowel is ?
$\begin{array}{1 1} 48 \\ 72 \\ 96 \\ 120 \end{array}$
jeemain
math
class11
ch7
permutations-and-combinations
easy
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asked
Sep 4, 2013
by
rvidyagovindarajan_1
edited
Apr 6, 2016
by
meena.p
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1 Answer
There are 4 consonants and 2 vowels.
No. of words having 3 alphabets = $^6C_3\times 3$
Out of these words $^4C_2\times 3!$ words have no vowels.
$\therefore$ No. of words which have at least one vowel
$=^6C_\times 3!-^4C_3\times 3!=3!(20-4)=96$
answered
Sep 4, 2013
by
rvidyagovindarajan_1
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