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6.Permutation and Combination
hard
The number of arrangements that can be formed from the letters $a, b, c, d, e,f$ taken $3$ at a time without repetition and each arrangement containing at least one vowel, is
A
$96$
B
$128$
C
$24$
D
$72$
(AIEEE-2012)
Solution
There are $2$ vowels and $4$ consonants in the letters $a,b,c,d,e,f.$
If we select one vowel, then number of arrangenment
${ = ^2}{C_1}{ \times ^4}{C_2} \times 3!$
$ = 2 \times \frac{{4 \times 3}}{2} \times 3 \times 2 = 72$
If we select two vowels, then number of arrangements
${ = ^2}{C_2}{ \times ^4}{C_1} \times 3!$
$ = 1 \times 4 \times 6 = 24$
Hence, total number of arrangemets
$=72+24=96$
Standard 11
Mathematics
