A selection committee is to be chosen consisting of 5 ex-technicians. Now there are 12 representatives from four zones. It has further been decided that if Mr. X is selected, Y and Z will not be selected and vice-versa. In how many ways it can be done?

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Answer = 12(C)5 - 3*9(C)4 - 6*9(C)3 - 9(C)2 12(C)5 represent the number of ways to select 5 things out of 12 things. 12(C)5 = total possible ways to select 5 out of 12 9(C)4 = total possible ways to select group such that only one from X,Y,Z must be there. so multiplied by 3 because 3 cases possible...
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Answer = 12(C)5 - 3*9(C)4 - 6*9(C)3 - 9(C)2 12(C)5 represent the number of ways to select 5 things out of 12 things. 12(C)5 = total possible ways to select 5 out of 12 9(C)4 = total possible ways to select group such that only one from X,Y,Z must be there. so multiplied by 3 because 3 cases possible (X, Y, Z)+ other 4 people from 12-3=9 person 9(C)3 = total possible ways to select group such that only two from X,Y,Z must be there. so multiplied by 3 because 3 cases possible (XY, YZ, ZX)+ other 3 people from 12-3=9 person. 9(C)2= all X,Y,Z must be there. (from 12-3 =9 we need to take 2 more people) Answer = Total cases - Not desired case read less
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10C5: when both are not included. 10C4: when one of them is included. Number of ways = 10C5 +10C4+ 10C4 = 672
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If X is chosen 9C4, If X is not Chosen 11C5=> 9C4+11C5 = 588
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