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Question Number 133424 by liberty last updated on 22/Feb/21
Given 10 white balls and ten black balls numbered 1,2,...,10. How many ways can we choose 6 balls such that (i) no two chosen balls have the same number (ii) two pairs of chosen balls have the same number?
$$\mathrm{Given}\:\mathrm{10}\:\mathrm{white}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{ten}\:\mathrm{black}\:\mathrm{balls} \\ $$$$\mathrm{numbered}\:\mathrm{1},\mathrm{2},…,\mathrm{10}.\:\mathrm{How}\:\mathrm{many}\: \\ $$$$\mathrm{ways}\:\mathrm{can}\:\mathrm{we}\:\mathrm{choose}\:\mathrm{6}\:\mathrm{balls}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{no}\:\mathrm{two}\:\mathrm{chosen}\:\mathrm{balls}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{number} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{two}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{chosen}\:\mathrm{balls}\:\mathrm{have} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{number}? \\ $$
Answered by EDWIN88 last updated on 22/Feb/21
(i) six number from the set {1,2,...,10} can be chosen (((10)),(( 6)) ) ways. Each of these numbers can appear in two ways on the chosen balls. There are (((10)),(( 6)) ). 2^6 = 13 440 ways of choosing 6 balls such that no two of them are denoted by the same number (ii) two pairs of balls having the same number can be chosen (((10)),(( 2)) ) ways. From the remaining 16 balls one can choose two balls numbered differently in ((8),(2) ) .2^2 ways. The answer in this case is (((10)),(( 2)) ) ((8),(2) ) .2^2 = 5 040
$$\left(\mathrm{i}\right)\:\mathrm{six}\:\mathrm{number}\:\mathrm{from}\:\mathrm{the}\:\mathrm{set}\:\left\{\mathrm{1},\mathrm{2},…,\mathrm{10}\right\}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{6}}\end{pmatrix}\:\mathrm{ways}.\:\mathrm{Each}\:\mathrm{of}\:\mathrm{these}\:\mathrm{numbers} \\ $$$$\mathrm{can}\:\mathrm{appear}\:\mathrm{in}\:\mathrm{two}\:\mathrm{ways}\:\mathrm{on}\:\mathrm{the}\:\mathrm{chosen}\:\mathrm{balls}. \\ $$$$\mathrm{There}\:\mathrm{are}\:\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{6}}\end{pmatrix}.\:\mathrm{2}^{\mathrm{6}} \:=\:\mathrm{13}\:\mathrm{440}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{choosing} \\ $$$$\mathrm{6}\:\mathrm{balls}\:\mathrm{such}\:\mathrm{that}\:\mathrm{no}\:\mathrm{two}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{denoted}\: \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{same}\:\mathrm{number} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{two}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{balls}\:\mathrm{having}\:\mathrm{the}\:\mathrm{same}\:\mathrm{number}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{chosen}\:\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\mathrm{ways}.\:\mathrm{From}\:\mathrm{the}\:\mathrm{remaining} \\ $$$$\mathrm{16}\:\mathrm{balls}\:\mathrm{one}\:\mathrm{can}\:\mathrm{choose}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{numbered} \\ $$$$\mathrm{differently}\:\mathrm{in}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:.\mathrm{2}^{\mathrm{2}} \:\mathrm{ways}.\:\mathrm{The}\:\mathrm{answer}\:\mathrm{in} \\ $$$$\mathrm{this}\:\mathrm{case}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:.\mathrm{2}^{\mathrm{2}} \:=\:\mathrm{5}\:\mathrm{040}\: \\ $$$$ \\ $$

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