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Question Number 124712 by benjo_mathlover last updated on 05/Dec/20

In how many ways  can 5 boys and 3 girls  be seated around a table if   (i) boy B_3  and G_2  are not adjacent  (ii) no girls are adjacent

$${In}\:{how}\:{many}\:{ways}\:\:{can}\:\mathrm{5}\:{boys}\:{and}\:\mathrm{3}\:{girls} \\ $$$${be}\:{seated}\:{around}\:{a}\:{table}\:{if}\: \\ $$$$\left({i}\right)\:{boy}\:{B}_{\mathrm{3}} \:{and}\:{G}_{\mathrm{2}} \:{are}\:{not}\:{adjacent} \\ $$$$\left({ii}\right)\:{no}\:{girls}\:{are}\:{adjacent}\: \\ $$

Answered by liberty last updated on 05/Dec/20

(i) The 5 boys and 2 girls not including   G_2  can be seated (7−1)! ways. G_2  has (7−2)  choices for a set not adjacent to B_3 . Thus  the desired number ways is 6!×5=3600  or the other method by Principle of   Complementation. If A is a subset of a  finite universal set U then ∣U \ A ∣ = ∣U∣−∣A∣  Now the number of ways to arrange the 5 boys  and 3 girls around a table so that B_3  and  G_2  are adjacent {B_3 ,G_2 } as an entity is  (7−1)!×2 = 1440  Thus the desired number of ways is   (8−1)!−1440=3600

$$\left({i}\right)\:{The}\:\mathrm{5}\:{boys}\:{and}\:\mathrm{2}\:{girls}\:{not}\:{including} \\ $$$$\:{G}_{\mathrm{2}} \:{can}\:{be}\:{seated}\:\left(\mathrm{7}−\mathrm{1}\right)!\:{ways}.\:{G}_{\mathrm{2}} \:{has}\:\left(\mathrm{7}−\mathrm{2}\right) \\ $$$${choices}\:{for}\:{a}\:{set}\:{not}\:{adjacent}\:{to}\:{B}_{\mathrm{3}} .\:{Thus} \\ $$$${the}\:{desired}\:{number}\:{ways}\:{is}\:\mathrm{6}!×\mathrm{5}=\mathrm{3600} \\ $$$${or}\:{the}\:{other}\:{method}\:{by}\:{Principle}\:{of}\: \\ $$$${Complementation}.\:{If}\:{A}\:{is}\:{a}\:{subset}\:{of}\:{a} \\ $$$${finite}\:{universal}\:{set}\:\mathcal{U}\:{then}\:\mid\mathcal{U}\:\backslash\:{A}\:\mid\:=\:\mid\mathcal{U}\mid−\mid{A}\mid \\ $$$${Now}\:{the}\:{number}\:{of}\:{ways}\:{to}\:{arrange}\:{the}\:\mathrm{5}\:{boys} \\ $$$${and}\:\mathrm{3}\:{girls}\:{around}\:{a}\:{table}\:{so}\:{that}\:{B}_{\mathrm{3}} \:{and} \\ $$$${G}_{\mathrm{2}} \:{are}\:{adjacent}\:\left\{{B}_{\mathrm{3}} ,{G}_{\mathrm{2}} \right\}\:{as}\:{an}\:{entity}\:{is} \\ $$$$\left(\mathrm{7}−\mathrm{1}\right)!×\mathrm{2}\:=\:\mathrm{1440} \\ $$$${Thus}\:{the}\:{desired}\:{number}\:{of}\:{ways}\:{is}\: \\ $$$$\left(\mathrm{8}−\mathrm{1}\right)!−\mathrm{1440}=\mathrm{3600} \\ $$

Commented by mr W last updated on 05/Dec/20

what did you get for (ii)?

$${what}\:{did}\:{you}\:{get}\:{for}\:\left({ii}\right)? \\ $$

Commented by liberty last updated on 06/Dec/20

for (ii) 4!×C_3 ^( 5) ×3! = 24×10×6 = 1440

$${for}\:\left({ii}\right)\:\mathrm{4}!×{C}_{\mathrm{3}} ^{\:\mathrm{5}} ×\mathrm{3}!\:=\:\mathrm{24}×\mathrm{10}×\mathrm{6}\:=\:\mathrm{1440} \\ $$

Answered by mr W last updated on 06/Dec/20

(i)  7!−2×6!=3600  (ii)  4!×P_3 ^5 =1440

$$\left({i}\right) \\ $$$$\mathrm{7}!−\mathrm{2}×\mathrm{6}!=\mathrm{3600} \\ $$$$\left({ii}\right) \\ $$$$\mathrm{4}!×{P}_{\mathrm{3}} ^{\mathrm{5}} =\mathrm{1440} \\ $$

Commented by benjo_mathlover last updated on 05/Dec/20

i think for (ii)   4!×5×4×3 = 24×60=1440  we first seat 5 boys around the   table (5−1)! =4! ways.  There are 5 ways to seat G_1  , as no  girls are adjacent , G_2  and G_3  have  4 and 3 choices respectively . Thus  the desired  number of ways  is 1440.

$${i}\:{think}\:{for}\:\left({ii}\right)\: \\ $$$$\mathrm{4}!×\mathrm{5}×\mathrm{4}×\mathrm{3}\:=\:\mathrm{24}×\mathrm{60}=\mathrm{1440} \\ $$$${we}\:{first}\:{seat}\:\mathrm{5}\:{boys}\:{around}\:{the}\: \\ $$$${table}\:\left(\mathrm{5}−\mathrm{1}\right)!\:=\mathrm{4}!\:{ways}. \\ $$$${There}\:{are}\:\mathrm{5}\:{ways}\:{to}\:{seat}\:{G}_{\mathrm{1}} \:,\:{as}\:{no} \\ $$$${girls}\:{are}\:{adjacent}\:,\:{G}_{\mathrm{2}} \:{and}\:{G}_{\mathrm{3}} \:{have} \\ $$$$\mathrm{4}\:{and}\:\mathrm{3}\:{choices}\:{respectively}\:.\:{Thus} \\ $$$${the}\:{desired}\:\:{number}\:{of}\:{ways} \\ $$$${is}\:\mathrm{1440}. \\ $$

Commented by liberty last updated on 06/Dec/20

4!×P_3 ^( 6)  is wrong. it should be 4!×P _3^5

$$\mathrm{4}!×{P}_{\mathrm{3}} ^{\:\mathrm{6}} \:{is}\:{wrong}.\:{it}\:{should}\:{be}\:\mathrm{4}!×{P}\:_{\mathrm{3}} ^{\mathrm{5}} \\ $$

Commented by mr W last updated on 06/Dec/20

thank you all sir!  certainly it is 4!P_3 ^5 . i had a typo.  to place the 5 boys there are 4! ways,  then to place the 3 girls in the 5   positions between the boys there are  P_3 ^5  ways. therefore totally 4!P_3 ^5  ways.

$${thank}\:{you}\:{all}\:{sir}! \\ $$$${certainly}\:{it}\:{is}\:\mathrm{4}!{P}_{\mathrm{3}} ^{\mathrm{5}} .\:{i}\:{had}\:{a}\:{typo}. \\ $$$${to}\:{place}\:{the}\:\mathrm{5}\:{boys}\:{there}\:{are}\:\mathrm{4}!\:{ways}, \\ $$$${then}\:{to}\:{place}\:{the}\:\mathrm{3}\:{girls}\:{in}\:{the}\:\mathrm{5}\: \\ $$$${positions}\:{between}\:{the}\:{boys}\:{there}\:{are} \\ $$$${P}_{\mathrm{3}} ^{\mathrm{5}} \:{ways}.\:{therefore}\:{totally}\:\mathrm{4}!{P}_{\mathrm{3}} ^{\mathrm{5}} \:{ways}. \\ $$

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