Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to
Solution
<p>$3 \text { Shelf empty: }(8,0,0,0) \rightarrow 1 \text { way }$</p>
<p>$$\left.2 \text { shelf empty: } \begin{array}{c}
(7,1,0,0) \\
(6,2,0,0) \\
(5,3,0,0) \\
(4,4,0,0)
\end{array}\right] \rightarrow 4 \text { ways }$$</p>
<p>$$\left.1 \text { shelf empty: } \begin{array}{cc}
(6,1,1,0) & (3,3,2,0) \\
(4,2,1,0) & (4,2,2,0) \\
(4,3,0) &
\end{array}\right] \rightarrow 5 \text { ways }$$</p>
<p>$$\left.0 \text { Shelf empty : } \begin{array}{ll}
(1,2,3,2) & (5,1,1,1) \\
(2,2,2,2) \\
(3,3,2,1,1) \\
(4,2,1,1) &
\end{array}\right] \rightarrow 5 \text { ways }$$</p>
<p>Total $=15$ ways</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
This question is part of PrepWiser's free JEE Main question bank. 135 more solved questions on Permutations and Combinations are available — start with the harder ones if your accuracy is >70%.