From a group of 12 students 8 are to be chosen for an excursion. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the 8 be chosen. ...Is it 9C5+9C8 or 9C5-9C8 Please clarify
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2You want to add the two cases where all 3 are included and where none of the 3 are included. – user84413 Jul 12 '16 at 02:42
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3$\binom{9}{5}$ counts the number of groups that have all the fussy ones, and $\binom{9}{8}$ counts the number of choices where they are left behind, as they should be. To get the total we add. – André Nicolas Jul 12 '16 at 02:44
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Okay I am convinced – Achari S Ganesha Jul 13 '16 at 16:45