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Let $x$ denote the number of ways of selecting at least one ball from a bag containing 3 identical red balls, 4 identical blue balls and 5 identical green balls. Let $y$ denote the number of ways in which a student will fail in an examination, when he has to write the examination in 5 different subjects. Then $x+y=$
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$150$
Number of ways to select at least one ball is
$\begin{aligned} x & =(3+1)(4+1)(5+1)-1 \\ & =4 \times 5 \times 6-1=119\end{aligned}$
and numbers of ways in which a student will fail in the examination is $y=2^5-1=31$
$\therefore \quad x+y=119+31=150$
$\begin{aligned} x & =(3+1)(4+1)(5+1)-1 \\ & =4 \times 5 \times 6-1=119\end{aligned}$
and numbers of ways in which a student will fail in the examination is $y=2^5-1=31$
$\therefore \quad x+y=119+31=150$
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