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How many 6 digit numbers can be formed using the digits 1,2,3,4 so that each of the digits should come atleast once in each number?

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Out of the 6 digits 4 digits are fixed.
In the remaining two digits, both the digits could be same
or both the digits could be different.
case (i): Both the digits are same.
No. 6 digit numbers =$^4C_1\times \large\frac{6!}{3!}$
(Both the digits could be any of the 4 digits..so .$4C_1$)
Case (ii) Both the digits are different.
The two digits could be any 2 of the 4 digits. so $^4C_2$
$\therefore$ The no. of 6 digit numbers in this case = $^4C_2\times\large\frac{6!}{2!.2!}$
$\therefore$ The required no. of 6 digit numbers=$^4C_1\times \large\frac{6!}{3!}$$+^4C_2\times \large\frac{6!}{2!.2!}$+
$=6!(\large\frac{2}{3}+\frac{3}{2})$$=6!\times\large\frac{13}{6}$$=1560$

 

answered Sep 4, 2013 by rvidyagovindarajan_1
 

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