Number of ways of arranging 8 identical books | vedclass

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(D) The problem asks for the number of ways to partition $n = 8$ identical items into $k = 4$ identical bins,where bins can be empty. This is equivale

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$15$

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(D) The problem asks for the number of ways to partition $n = 8$ identical items into $k = 4$ identical bins,where bins can be empty. This is equivalent to finding the number of partitions of $8$ into at most $4$ parts,denoted as $p_4(8)$.We list the partitions of $8$ into at most $4$ parts:$1$ part: $(8) \rightarrow 1$ way$2$ parts: $(7,1), (6,2), (5,3), (4,4) \rightarrow 4$ ways$3$ parts: $(6,1,1), (5,2,1), (4,3,1), (4,2,2), (3,3,2) \rightarrow 5$ ways$4$ parts: $(5,1,1,1), (4,2,1,1), (3,3,1,1), (3,2,2,1), (2,2,2,2) \rightarrow 5$ waysTotal number of ways = $1 + 4 + 5 + 5 = 15$.

Number of ways of arranging $8$ identical books into $4$ identical shelves where any number of shelves may remain empty is equal to

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