Category: Permutation and Combination

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…

Question Number 135303 by bobhans last updated on 12/Mar/21 $$$$in how many ways can 10 people be seated in a round table if…

Question Number 135239 by benjo_mathlover last updated on 11/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 4105 by prakash jain last updated on 28/Dec/15 $$\mathrm{How}\mathrm{many}\mathrm{distinct}\mathrm{ways}\mathrm{are}\mathrm{there}\mathrm{for}$$$$\mathrm{a}\mathrm{knight}\mathrm{to}\mathrm{reach}\mathrm{from}\mathrm{bottom}\mathrm{left}\mathrm{corner}\mathrm{of}$$$$\mathrm{chessboard}\mathrm{to}\mathrm{top}\mathrm{right}\mathrm{corner}.$$$$\left(\mathrm{knight}\mathrm{going}\mathrm{from}\mathrm{square}\mathrm{a}1\mathrm{to}\mathrm{h}8\right).$$ Commented by Filup last updated on…