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Question Number 63490 by kumaranjul306@gmail.com last updated on 04/Jul/19

$$\mathrm{A}\mathrm{father}\mathrm{with}8\mathrm{children}\mathrm{takes}3\mathrm{at}\mathrm{a}$$$$\mathrm{time}\mathrm{to}\mathrm{the}\mathrm{garden}\mathrm{as}\mathrm{often}\mathrm{as}\mathrm{he}$$$$\mathrm{without}\mathrm{taking}\mathrm{the}\mathrm{same}3\mathrm{children}$$$$\mathrm{together}\mathrm{more}\mathrm{than}\mathrm{once}.\mathrm{The}\mathrm{number}$$$$\mathrm{of}\mathrm{times}\mathrm{he}\mathrm{will}\mathrm{go}\mathrm{to}\mathrm{the}\mathrm{garden}\mathrm{is}$$

Commented by Prithwish sen last updated on 05/Jul/19

$$\mathrm{out}\mathrm{of}8\mathrm{taking}3\mathrm{at}\mathrm{a}\mathrm{time},\mathrm{the}\mathrm{number}\mathrm{of}$$$$\mathrm{ways}\mathrm{will}\mathrm{be}{\mathrm{C}}_3^8=56\mathrm{ways}$$$$\mathrm{Now}\mathrm{out}\mathrm{of}\mathrm{remaining}5\mathrm{taking}3\mathrm{at}\mathrm{a}\mathrm{time}$$$$\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ways}{\mathrm{C}}_3^5=10\mathrm{ways}$$$$\mathrm{and}\mathrm{the}\mathrm{remaining}2\mathrm{will}\mathrm{be}\mathrm{taken}\mathrm{in}1\mathrm{way}$$$$\therefore \mathrm{the}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{ways}=56+10+1=67$$$$\mathrm{please}\mathrm{check}.$$

Answered by Rio Michael last updated on 05/Jul/19

$$ifhetakesthefirst3childrenin2!ways=numberoftimes$$$$hewillgoleaving2childrenbehind=2times$$$$hecomesbackwithasinglechildandgoes1!=1time$$$$thenumberoftimeshegoestothatgarden=3timesforafirstselection$$$$then3\times 6times=18timesforatotalselection.$$

Answered by ajfour last updated on 05/Jul/19

$${}^8{C}_3=\frac{6\times 7\times 8}{2\times 3}=56.$$

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