A is the answer???

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We will have to take all cases in which we select one "indistinguishable" item, 2 "indistinguishable" item and so on till n "instinguishabe item.

Let indistinguishable item be k-set items, distinct items be s-set items.( or vice versa it will not change answer)

When we select 0 item from set of k-items then we can have n from s-items, when 1 item from set of k-items then n-1 from s-set and so on......

So total ways is,

^{n}C_{}_{0 }+ ^{n}C_{1 }+ _{}.............+^{n}C_{n }= 2^{n }(binomial)

http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/chinonyerem1.html