Question 1174603
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1st patient occupies any of          10 beds   (10 options)

2nd patient occupies any of remaining 9 beds   (9 options)

3rd patient occupies any of remaining 8 beds   (8 options)

4th patient occupies any of remaining 7 beds   (7 options)

5th patient occupies any of remaining 6 beds   (6 options)



THEREFORE, there are  10*9*8*7*6 = 30240 different ways to accommodate.     <U>ANSWER</U>
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Solved.


The product of 5 sequential integer numbers in descending order, starting from the number 10.