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Question Number 43031 by aseerimad last updated on 06/Sep/18
Commented by MrW3 last updated on 06/Sep/18
$$\mathrm{8}!×\left(\mathrm{9}+{C}_{\mathrm{2}} ^{\mathrm{9}} \right)=\mathrm{8}!×\mathrm{45}=\mathrm{5}×\mathrm{9}!=\mathrm{1814400} \\ $$
Commented by MrW3 last updated on 07/Sep/18
![Explanation: We arrange at first the other 8 speakers, there are 8! ways to do this. There are two methods to arrange the speakers A and B then: method 1: A and B are next to each other. There are 9 possibilities. E.g. XXXXXXBAXX, or XXXXXXXXBA orBAXXXXXXXX. method 2: A and B are separated by at least one of the other 8 people. We can place A and B in 9 possible positions. There are C_2 ^9 ways. E.g. XXBXXXXAXX, or XXXXXXXBXA or BXXXXXXXXA. Thus the result is 8!×(9+C_2 ^9 )=8!×45=5×9! Generally in case of n people: (n−2)!×[(n−1)+C_2 ^(n−1) ] =(n−2)!×[(n−1)+(((n−1)(n−2))/2)] =((n(n−1)(n−2)!)/2) =((n!)/2)](Q43056.png)
$${Explanation}: \\ $$$${We}\:{arrange}\:{at}\:{first}\:{the}\:{other}\:\mathrm{8}\:{speakers}, \\ $$$${there}\:{are}\:\mathrm{8}!\:{ways}\:{to}\:{do}\:{this}. \\ $$$${There}\:{are}\:{two}\:{methods}\:{to}\:{arrange}\:{the}\: \\ $$$${speakers}\:{A}\:{and}\:{B}\:{then}: \\ $$$${method}\:\mathrm{1}:\:{A}\:{and}\:{B}\:{are}\:{next}\:{to}\:{each}\:{other}.\:{There} \\ $$$${are}\:\mathrm{9}\:{possibilities}.\:{E}.{g}.\:{XXXXXXBAXX}, \\ $$$${or}\:{XXXXXXXXBA}\:{orBAXXXXXXXX}. \\ $$$${method}\:\mathrm{2}:\:{A}\:{and}\:{B}\:{are}\:{separated}\:{by}\:{at} \\ $$$${least}\:{one}\:{of}\:{the}\:{other}\:\mathrm{8}\:{people}.\:{We} \\ $$$${can}\:{place}\:{A}\:{and}\:{B}\:{in}\:\mathrm{9}\:{possible}\:{positions}. \\ $$$${There}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{9}} \:{ways}.\:{E}.{g}.\:{XXBXXXXAXX}, \\ $$$${or}\:{XXXXXXXBXA}\:{or}\:{BXXXXXXXXA}. \\ $$$${Thus}\:{the}\:{result}\:{is} \\ $$$$\mathrm{8}!×\left(\mathrm{9}+{C}_{\mathrm{2}} ^{\mathrm{9}} \right)=\mathrm{8}!×\mathrm{45}=\mathrm{5}×\mathrm{9}! \\ $$$$ \\ $$$${Generally}\:{in}\:{case}\:{of}\:{n}\:{people}: \\ $$$$\left({n}−\mathrm{2}\right)!×\left[\left({n}−\mathrm{1}\right)+{C}_{\mathrm{2}} ^{{n}−\mathrm{1}} \right] \\ $$$$=\left({n}−\mathrm{2}\right)!×\left[\left({n}−\mathrm{1}\right)+\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{2}}\right] \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)!}{\mathrm{2}} \\ $$$$=\frac{{n}!}{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
$${arrangement}\:{of}\:{speaker}\:{like} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\left.{supose}\:\mathrm{1}\right){B}\:{is}\:{first}\:{speaker}\:{then}\:{A}\:{can}\:{be}\:{any} \\ $$$${position}\:{out}\:{of}\:{remaining}\:\mathrm{9}\:{so} \\ $$$${permutation}\:{is}\:\mathrm{1}×\mathrm{9}×\mathrm{8}! \\ $$$${here}\:\mathrm{1}\:{for}\:{B}\:{and}\:\:\mathrm{8}!\:{remaining}\:\mathrm{8}{speaker} \\ $$$$\left.\mathrm{2}\right){when}\:{B}\:{second}\:{speaker}\:{A}\:{can}\:{be}\:\mathrm{3}{rd}\:\mathrm{4}{th}...\mathrm{10}{th} \\ $$$${so}\:\mathrm{1}×\mathrm{8}×\mathrm{8}! \\ $$$$\left.\mathrm{3}\right){when}\:{B}\:{third}\:{speaker}\:{A}\:\:{can}\:{be}\:\mathrm{4}{th}\:\mathrm{5}{th}...\mathrm{10}{th} \\ $$$$\mathrm{1}×\mathrm{7}×\mathrm{8}! \\ $$$$\left.\mathrm{4}\right)\:{when}\:{B}\:\mathrm{4}{th}\:{speaker}\:{A}\:{can}\:{be}\:\mathrm{5}{th}\:\mathrm{6}{th}...\mathrm{10}{th} \\ $$$$\mathrm{1}×\mathrm{6}×\mathrm{8}! \\ $$$$\left.\mathrm{5}\right){when}\:{B}\:\mathrm{5}{th}\:{soeaker}\:{A}\:{can}\:{be}\:\mathrm{6}{th}\:\mathrm{7}{th}...\mathrm{10}{th}=\mathrm{1} \\ $$$$\mathrm{1}×\mathrm{5}×\mathrm{8}! \\ $$$$\left.\mathrm{6}\right){when}\:{B}\:\mathrm{6}{th}\:{speaker}\:{A}\:{can}\:{be}\:\mathrm{7}{th}\:\mathrm{8}{th}..\mathrm{10}{th} \\ $$$$\mathrm{1}×\mathrm{4}×\mathrm{8}! \\ $$$$\left.\mathrm{7}\right){when}\:{B}\:\mathrm{7}{th}\:{speaker}\:{A}\:{can}\:{be}\:\mathrm{8}{th}\:\mathrm{9}{th}\:\mathrm{10}{th} \\ $$$$\mathrm{1}×\mathrm{3}×\mathrm{8}! \\ $$$$\left.\mathrm{8}\right){when}\:{B}\:\mathrm{8}{th}\:{speaker}\:{A}\left[{can}\:{be}\left[\mathrm{9}{th}\:\mathrm{10}{th}\right.\right. \\ $$$$\mathrm{1}×\mathrm{2}×\mathrm{8}! \\ $$$$\left.\mathrm{9}\right){when}\:{B}\:\mathrm{9}{th}\:{speaker}\:{A}\:{can}\:{be}\:\mathrm{10}{th} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{8}! \\ $$$${so}\:{ans}\:{is}\: \\ $$$$\mathrm{1}×\mathrm{8}!\left(\mathrm{9}+\mathrm{8}+\mathrm{7}+\mathrm{6}+\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}\right) \\ $$$$=\mathrm{1}×\mathrm{8}!×\frac{\mathrm{9}}{\mathrm{2}}\left\{\mathrm{2}×\mathrm{9}+\left(\mathrm{9}−\mathrm{1}\right)×−\mathrm{1}\right\} \\ $$$$=\mathrm{1}×\mathrm{8}!×\frac{\mathrm{9}}{\mathrm{2}}\left(\mathrm{18}−\mathrm{8}\right) \\ $$$$=\mathrm{8}!×\mathrm{9}×\mathrm{5} \\ $$$$=\mathrm{9}!×\mathrm{5} \\ $$
Answered by MrW3 last updated on 06/Sep/18
$${Here}\:{is}\:{an}\:{easy}\:{solution}: \\ $$$${To}\:{arrange}\:\mathrm{10}\:{people}\:{there}\:{are}\:\mathrm{10}!\:{ways}. \\ $$$${The}\:{number}\:{of}\:{arrangements}\:{with} \\ $$$${A}\:{before}\:{B}\:{is}\:{the}\:{same}\:{as}\:{the}\:{number} \\ $$$${of}\:{arrangements}\:{with}\:{A}\:{behind}\:{B}.\: \\ $$$${Thus}\:{the}\:{solution}\:{is}\:\frac{\mathrm{10}!}{\mathrm{2}}=\mathrm{1814400}. \\ $$
Commented by ajfour last updated on 07/Sep/18
$${excellent}\:{Sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18
$${yes}\:{sir}...{thank}\:{you}... \\ $$
Commented by malwaan last updated on 07/Sep/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$