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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?
Given10whiteballsandtenblackballsnumbered1,2,,10.Howmanywayscanwechoose6ballssuchthat(i)notwochosenballshavethesamenumber(ii)twopairsofchosenballshavethesamenumber?
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
(i)sixnumberfromtheset{1,2,,10}canbechosen(106)ways.Eachofthesenumberscanappearintwowaysonthechosenballs.Thereare(106).26=13440waysofchoosing6ballssuchthatnotwoofthemaredenotedbythesamenumber(ii)twopairsofballshavingthesamenumbercanbechosen(102)ways.Fromtheremaining16ballsonecanchoosetwoballsnumbereddifferentlyin(82).22ways.Theanswerinthiscaseis(102)(82).22=5040

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