Question #325619

Suppose that an operating room needs to handle three knee, four hip, and five shoulder surgeries.

⦁ How many different sequences are possible?

⦁ How many different sequences have all hip, knee, and shoulder surgeries scheduled consecutively?

⦁ How many different schedules begin and end with a knee surgery?


1
Expert's answer
2022-04-14T08:26:51-0400

1:(3+4+5)!=12!=4.79002×1082:3!=6waystoarrangeasequenceofthreetypes3!=6toarrangeasequenceofkneesurgeries4!=24toarrangeasequenceofhipsurgeries5!=120toarrangeasequenceofshouldersurgeriesThetotalis6624120=103680ways3:32=6waystochoosethefirstandthelastkneesurgery(1+4+5)!=10!=3.6288×106waystoarrangetherestsurgeriesThetotalis63.6288×106=2.17728×107ways1:\\\left( 3+4+5 \right) !=12!=4.79002\times 10^8\\2:\\3!=6\,\,ways\,\,to\,\,arrange\,\,a\,\,sequence\,\,of\,\,three\,\,types\\3!=6\,\,to\,\,arrange\,\,a\,\,sequence\,\,of\,\,knee\,\,surgeries\\4!=24\,\,to\,\,arrange\,\,a\,\,sequence\,\,of\,\,hip\,\,surgeries\\5!=120\,\,to\,\,arrange\,\,a\,\,sequence\,\,of\,\,shoulder\,\,surgeries\\The\,\,total\,\,is\,\,\\6\cdot 6\cdot 24\cdot 120=103680\,\,ways\\3:\\3\cdot 2=6\,\,ways\,\,to\,\,choose\,\,the\,\,first\,\,and\,\,the\,\,last\,\,knee\,\,surgery\\\left( 1+4+5 \right) !=10!=3.6288\times 10^6\,\,ways\,\,to\,\,arrange\,\,the\,\,rest\,\,surgeries\\The\,\,total\,\,is\\6\cdot 3.6288\times 10^6=2.17728\times 10^7\,\,ways\,\,\\


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS