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Permutations and Combinations
Total number of ways in which 15 identical blanks can be distributed among 4 persons so that each of them gets atleast two blankets is equal to
$\begin{array}{1 1}(A)\;10C_3\\(B)\;9C_3\\(C)\;11C_3\\(D)\;\text{None of these}\end{array} $
cbse
math
class11
ch7
permutations-and-combinations
additionalproblem
q27
sec-b
medium
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asked
Jun 20, 2014
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sreemathi.v
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1 Answer
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$C(n,r)=\large\frac{n!}{r!(n-r)!}$
$x_1+x_2+x_3+x_4=15$ and $x_i \geq 2$
$\Rightarrow (x_1-2)+(x_2-2)+(x_3-2)+(x_4-2)=7$
$\Rightarrow y_1+y_2+y_3+y_4=7$
Where $y_i=x_i-2 \geq 0$
$\Rightarrow 10C_7=10C_3$ are the number of non-negative.
Hence (A) is the correct answer.
answered
Jun 20, 2014
by
sreemathi.v
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