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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only is

$\begin{array}{1 1}(A)\;55\\(B)\;66\\(C)\;77\\(D)\;88\end{array} $

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  • $n!=n(n-1)(n-2)(n-3).....(1)(2)(3)$
There are two possible cases.
Case I:Five 1's one 2's one 3's
Number of numbers =$\large\frac{7!}{5!}$
$\Rightarrow \large\frac{7\times 6\times 5!}{5!}$$=42$
Case II: Four 1's three 2's
Number of numbers =$\large\frac{7!}{4!3!}$
$\Rightarrow \large\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}$$=35$
Total number of numbers =42+35=77
Hence (C) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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