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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 last updated on 05/Feb/21
![more generally with b elements of one kind and m of another randomly arranged in a line , the expected number of unlike adjacents elements is (m+b−1) [ ((bm)/((m+b)(m+b−1)))+((mb)/((m+b)(m+b−1))) ]= ((2mb)/(m+b)) . In our question b=8 ,m=7 giving ((2.8.7)/(15))=7(7/(15))](Q131462.png)
$${more}\:{generally}\:{with}\:{b}\:{elements} \\ $$$${of}\:{one}\:{kind}\:{and}\:{m}\:{of}\:{another}\: \\ $$$${randomly}\:{arranged}\:{in}\:{a}\:{line}\:,\:{the}\: \\ $$$${expected}\:{number}\:{of}\:{unlike} \\ $$$${adjacents}\:{elements}\:{is}\: \\ $$$$\left({m}+{b}−\mathrm{1}\right)\:\left[\:\frac{{bm}}{\left({m}+{b}\right)\left({m}+{b}−\mathrm{1}\right)}+\frac{{mb}}{\left({m}+{b}\right)\left({m}+{b}−\mathrm{1}\right)}\:\right]= \\ $$$$\:\frac{\mathrm{2}{mb}}{{m}+{b}}\:.\:{In}\:{our}\:{question}\:{b}=\mathrm{8}\:,{m}=\mathrm{7} \\ $$$${giving}\:\frac{\mathrm{2}.\mathrm{8}.\mathrm{7}}{\mathrm{15}}=\mathrm{7}\frac{\mathrm{7}}{\mathrm{15}} \\ $$