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Re: A chef has 10 spices available to season soups and uses no fewer than
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03 Feb 2023, 22:14
Given that a Out of 10 spices, the Chef chooses no fewer than 5 and no more than 8 of the spices to season his vegetable soup.
Now the number of species which we can choose can be between 5 and 8(both inclusive) => 5, 6, 7, 8
If we are choosing 5, then we can choose 5 out of 10 on 10C5 ways = \(\frac{10!}{(5!)*(10-5)!}\) = \(\frac{10!}{5!*5!}\) = \(\frac{10*9*8*7*6*5!}{5!*1*2*3*4*5}\) = 252 ways
If we are choosing 6, then we can choose 6 out of 10 on 10C6 ways = \(\frac{10!}{(5!)*(10-6)!}\) = \(\frac{10!}{6!*4!}\) = \(\frac{10*9*8*7*6!}{6!*1*2*3*4}\) = 210 ways
If we are choosing 7, then we can choose 7 out of 10 on 10C7 ways = \(\frac{10!}{(7!)*(10-7)!}\) = \(\frac{10!}{7!*3!}\) = \(\frac{10*9*8*7!}{7!*1*2*3}\) = 120 ways
If we are choosing 8, then we can choose 8 out of 10 on 10C8 ways = \(\frac{10!}{(8!)*(10-8)!}\) = \(\frac{10!}{8!*2!}\) = \(\frac{10*9*8!}{8!*1*2}\) = 45 ways
Quantity A: The number of spices that produces the greatest number of ways to season the vegetable soup
Greatest number is produced when we take number of spices as 5
=> Quantity A = 5
Quantity B: The number of spices that produces the least number of ways to season the vegetable soup
Least number is produced when we take number of spices as 8
=> Quantity B = 8
Clearly, Quantity A < Quantity B
So, Answer will be B
Hope it helps!