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Category: Permutation and Combination

determinant-old-plate-EEU-874-new-plate-1BXK-267-Old-California-license-plate-consisted-of-a-sequence-of-three-letters-followed-by-three-digits-see-figure-above-Assuming-that-any-s

Question Number 131464 by bemath last updated on 05/Feb/21 $$\:\begin{array}{|c|c|}{{old}\:{plate}\::\:{EEU}\:\mathrm{874}}\\{{new}\:{plate}\::\:\mathrm{1}{BXK}\:\mathrm{267}}\\\hline\end{array} \\ $$$${Old}\:{California}\:{license}\:{plate}\: \\ $$$${consisted}\:{of}\:{a}\:{sequence}\:{of} \\ $$$${three}\:{letters}\:{followed}\:{by}\:{three} \\ $$$${digits}\:\left({see}\:{figure}\:{above}\right). \\ $$$${Assuming}\:{that}\:{any}\:{sequence} \\ $$$${of}\:{letters}\:{and}\:{digits}\:{was}\:{allowed} \\ $$$$\left({though}\:{actually}\:{some}\:{combinations}\right. \\…

Eight-eligible-bachelor-and-seven-beautiful-models-happen-randomly-to-have-purchased-single-seats-in-the-same-15-seats-row-of-theather-On-the-average-how-many-pairs-of-adjacent-seats-are-ticketed-

Question Number 131459 by bramlexs22 last updated on 05/Feb/21 $${Eight}\:{eligible}\:{bachelor}\:{and}\:{seven}\:{beautiful} \\ $$$${models}\:{happen}\:{randomly}\:{to}\:{have}\: \\ $$$${purchased}\:{single}\:{seats}\:{in}\:{the}\:{same}\:\mathrm{15}−{seats} \\ $$$${row}\:{of}\:{theather}.\:{On}\:{the}\:{average}\:,\:{how}\:{many} \\ $$$${pairs}\:{of}\:{adjacent}\:{seats}\:{are}\:{ticketed} \\ $$$${for}\:{marriageable}\:{couples}\:? \\ $$ Answered by bemath…

If-and-are-the-coefficient-of-x-8-and-x-24-respectively-in-the-expansion-of-x-4-2-1-x-4-10-in-powers-of-x-then-is-equal-to-

Question Number 131248 by bramlexs22 last updated on 03/Feb/21 $${If}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{coefficient}\: \\ $$$${of}\:{x}^{\mathrm{8}} \:{and}\:{x}^{−\mathrm{24}} \:{respectively}\: \\ $$$${in}\:{the}\:{expansion}\:{of}\:\left[\:{x}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\right]^{\mathrm{10}} \\ $$$${in}\:{powers}\:{of}\:{x}\:{then}\:\frac{\alpha}{\beta}\:{is}\:{equal}\:{to}\: \\ $$ Answered by EDWIN88…

Five-children-sitting-one-behind-the-other-in-a-five-seater-merry-go-round-decide-to-switch-seats-so-that-each-child-has-new-companion-in-front-In-how-many-ways-can-this-be-done-

Question Number 131162 by EDWIN88 last updated on 02/Feb/21 $${Five}\:{children}\:{sitting}\:{one}\:{behind} \\ $$$${the}\:{other}\:{in}\:{a}\:{five}\:{seater}\:{merry}−{go}−{round} \\ $$$$,{decide}\:{to}\:{switch}\:{seats}\:{so}\:{that}\:{each} \\ $$$${child}\:{has}\:{new}\:{companion}\:{in}\:{front}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{this}\:{be}\:{done}? \\ $$ Commented by mr W last…

Sum-the-series-n-0-P-r-n-x-n-n-where-P-r-n-is-a-polynomial-of-degree-r-in-n-

Question Number 23 by user1 last updated on 25/Jan/15 $$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{P}}_{{r}} \left({n}\right)\frac{{x}^{{n}} }{{n}!}\:,\:\mathrm{where}\:\boldsymbol{\mathrm{P}}_{{r}} \left({n}\right)\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{of}\:\mathrm{degree}\:{r}\:\mathrm{in}\:{n}. \\ $$ Answered by user1 last updated on…

Prove-that-lim-n-2n-2n-1-ln-n-p-0-n-ln-1-p-2-ln-e-pi-e-pi-

Question Number 143740 by Willson last updated on 17/Jun/21 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}2n}−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{n}\right)+\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\:\mathrm{ln}\left({e}^{\pi} −{e}^{−\pi} \right) \\ $$ Answered by TheHoneyCat last updated…

In-a-basic-version-of-pocker-each-players-is-dealth-5-cards-from-a-standard-52-cards-no-pocker-How-many-diferent-5-cards-pocker-hand-are-there-

Question Number 12590 by tawa last updated on 26/Apr/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{basic}\:\mathrm{version}\:\mathrm{of}\:\mathrm{pocker}\:,\:\mathrm{each}\:\mathrm{players}\:\mathrm{is}\:\mathrm{dealth}\:\mathrm{5}\:\mathrm{cards}\:\mathrm{from}\:\mathrm{a}\:\mathrm{standard} \\ $$$$\mathrm{52}\:\mathrm{cards}\:\left(\mathrm{no}\:\mathrm{pocker}\right).\:\mathrm{How}\:\mathrm{many}\:\mathrm{diferent}\:\mathrm{5}\:\mathrm{cards}\:\mathrm{pocker}\:\mathrm{hand}\:\mathrm{are}\:\mathrm{there}\:??? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-143627

Question Number 143627 by bobhans last updated on 16/Jun/21 Answered by Olaf_Thorendsen last updated on 16/Jun/21 $$\mathrm{X}\:=\:\frac{\left(\mathrm{3}^{{x}} −\mathrm{4}^{{x}} \right)\left(\mathrm{3}^{{x}} −\mathrm{4}.\mathrm{4}^{{x}} \right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{6}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{4}}{\mathrm{3}}\right)\right)} \\ $$$${etc}… \\…