Category: Permutation and Combination

Question Number 5444 by 3 last updated on 15/May/16 $$8$$ Answered by FilupSmith last updated on 15/May/16 $$=7+1$$ Answered by Rasheed…

Question Number 136417 by I want to learn more last updated on 21/Mar/21 Answered by EDWIN88 last updated on 22/Mar/21 $$\mathrm{even}\mathrm{number}=\{2,4\},\mathrm{odd}\mathrm{number}=\{1,3,5\}$$$$\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{on}\mathrm{two}\mathrm{balls}$$$$\mathrm{have}\:\mathrm{probabability}\:=\:\frac{\mathrm{1}+\mathrm{3}}{\mathrm{C}_{\mathrm{2}}…

Question Number 5280 by Rasheed Soomro last updated on 04/May/16 $$\mathrm{A}\mathrm{delegation}\mathrm{of}4\mathrm{people}\mathrm{is}\mathrm{to}\mathrm{be}$$$$\mathrm{selected}\mathrm{from}5\mathrm{women}\mathrm{and}6\mathrm{men}.$$$$\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{delegations}\mathrm{if}$$$$\left(\mathbf{a}\right)\mathrm{there}\mathrm{are}\mathrm{no}\mathrm{restrictions},$$$$\left(\mathbf{b}\right)\mathrm{there}\mathrm{is}\mathrm{at}\mathrm{least}1\mathrm{woman},$$$$\left(\mathbf{c}\right)\mathrm{there}\mathrm{are}\mathrm{at}\mathrm{least}2\mathrm{women}.$$$$\mathrm{One}\mathrm{of}\mathrm{the}\mathrm{men}\mathrm{cannot}\mathrm{get}\mathrm{along}\mathrm{with}$$$$\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{women}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}…

Question Number 5254 by Rasheed Soomro last updated on 02/May/16 $$\mathrm{Calculate}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{in}\mathrm{which}$$$$\left(\mathbf{a}\right)5\mathrm{children}\mathrm{can}\mathrm{be}\mathrm{divided}\mathrm{into}\mathrm{groups}$$$$\mathrm{of}2\mathrm{and}3,$$$$\left(\mathbf{b}\right)9\mathrm{children}\mathrm{can}\mathrm{be}\mathrm{divided}\mathrm{into}\mathrm{groups}$$$$\mathrm{of}5\mathrm{and}4,$$$$\mathrm{Hence}\mathrm{calculate}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{in}$$$$\mathrm{which}9\mathrm{children}\mathrm{can}\mathrm{be}\mathrm{divided}\mathrm{into}$$$$\mathrm{groups}\:\mathrm{of}\:\mathrm{2},\mathrm{3}\:\mathrm{and}\:\mathrm{4}.…

Question Number 135599 by liberty last updated on 14/Mar/21 Answered by mathDivergent last updated on 14/Mar/21 $$2^8,\mathrm{I}\mathrm{have}\mathrm{blunder}$$ Commented by EDWIN88 last updated…