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Question Number 200051 by cortano12 last updated on 13/Nov/23

There are many ways to arrange 3 red balls and 9 black balls in a circle so that there are a minimum of 2 black balls between 2 adjacent red balls. (a) 180×8! (b) 240×7! (c) 364×6! (d) 282×4! (e) 144×5!

$$ \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{arrange}\:\mathrm{3}\:\mathrm{red} \\ $$$$\:\mathrm{balls}\:\mathrm{and}\:\mathrm{9}\:\mathrm{black}\:\mathrm{balls}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\: \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{a}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{2} \\ $$$$\mathrm{black}\:\mathrm{balls}\:\mathrm{between}\:\mathrm{2}\:\mathrm{adjacent}\:\mathrm{red} \\ $$$$\mathrm{balls}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{180}×\mathrm{8}!\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{240}×\mathrm{7}!\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{364}×\mathrm{6}! \\ $$$$\:\left(\mathrm{d}\right)\:\mathrm{282}×\mathrm{4}!\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{144}×\mathrm{5}!\: \\ $$

Commented by mr W last updated on 13/Nov/23

different or identical balls?

$${different}\:{or}\:{identical}\:{balls}? \\ $$

Commented by cortano12 last updated on 13/Nov/23

different balls

$$\mathrm{different}\:\mathrm{balls} \\ $$

Answered by mr W last updated on 13/Nov/23

to arrange the red balls: 2!=2 to arrange the black balls: 2+2+5 ⇒((3!)/(2!))×2!×2!×5!=1440 2+3+4 ⇒3!×2!×3!×4!=1728 3+3+3 ⇒3!×3!×3!=216 ⇒2×(1440+1728+216)=6768 ✓ ⇒answer (d)

$${to}\:{arrange}\:{the}\:{red}\:{balls}:\:\mathrm{2}!=\mathrm{2} \\ $$$${to}\:{arrange}\:{the}\:{black}\:{balls}: \\ $$$$\mathrm{2}+\mathrm{2}+\mathrm{5}\:\Rightarrow\frac{\mathrm{3}!}{\mathrm{2}!}×\mathrm{2}!×\mathrm{2}!×\mathrm{5}!=\mathrm{1440} \\ $$$$\mathrm{2}+\mathrm{3}+\mathrm{4}\:\Rightarrow\mathrm{3}!×\mathrm{2}!×\mathrm{3}!×\mathrm{4}!=\mathrm{1728} \\ $$$$\mathrm{3}+\mathrm{3}+\mathrm{3}\:\Rightarrow\mathrm{3}!×\mathrm{3}!×\mathrm{3}!=\mathrm{216} \\ $$$$\Rightarrow\mathrm{2}×\left(\mathrm{1440}+\mathrm{1728}+\mathrm{216}\right)=\mathrm{6768}\:\checkmark \\ $$$$\Rightarrow{answer}\:\left({d}\right) \\ $$

Commented by cortano12 last updated on 13/Nov/23

minimum 2 red balls between two adjacent black balls sir. it′s possible 3 red balls between two adjacent black balls?

$$\mathrm{minimum}\:\mathrm{2}\:\mathrm{red}\:\mathrm{balls}\:\mathrm{between}\:\mathrm{two}\: \\ $$$$\mathrm{adjacent}\:\mathrm{black}\:\mathrm{balls}\:\mathrm{sir}. \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{3}\:\mathrm{red}\:\mathrm{balls}\:\mathrm{between}\:\mathrm{two} \\ $$$$\:\mathrm{adjacent}\:\mathrm{black}\:\mathrm{balls}? \\ $$

Commented by mr W last updated on 13/Nov/23

no, 2, 3, 4 or 5 black balls between two red balls as the question requests.

$${no},\:\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:{or}\:\mathrm{5}\:\boldsymbol{{black}}\:{balls}\:{between} \\ $$$${two}\:\boldsymbol{{red}}\:{balls}\:{as}\:{the}\:{question}\:{requests}. \\ $$

Commented by cortano12 last updated on 13/Nov/23

l got the answer A

$$\mathrm{l}\:\mathrm{got}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{A} \\ $$

Commented by cortano12 last updated on 13/Nov/23

Commented by mr W last updated on 13/Nov/23

yes, you are right! (A) is correct.

$${yes},\:{you}\:{are}\:{right}!\:\left({A}\right)\:{is}\:{correct}. \\ $$

Answered by MM42 last updated on 13/Nov/23

for“ red” → 2 for “black” 5,2,2→3 / 4,3,2→6 / 3,3,3→1⇒10 ans=2×10×9!=180×8! ✓

$${for}``\:{red}''\:\rightarrow\:\mathrm{2} \\ $$$${for}\:``{black}'' \\ $$$$\mathrm{5},\mathrm{2},\mathrm{2}\rightarrow\mathrm{3}\:\:/\:\:\mathrm{4},\mathrm{3},\mathrm{2}\rightarrow\mathrm{6}\:\:/\:\mathrm{3},\mathrm{3},\mathrm{3}\rightarrow\mathrm{1}\Rightarrow\mathrm{10} \\ $$$${ans}=\mathrm{2}×\mathrm{10}×\mathrm{9}!=\mathrm{180}×\mathrm{8}!\:\:\checkmark \\ $$$$ \\ $$

Commented by MM42 last updated on 13/Nov/23

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