no fewer than 5

which means 5 but also 6,7,8

no more than 8

which means 8,7/6

I.E

our x is between 5 and 8 included

\(5\leq x < \leq 8\)

I'm not arguing with that, but I still don't understand how the answer is A.

We get the greatest number of ways when we use 5 spices - Quantity A.
We get the least number of ways when we use 8 spices - Quantity B.

Quantity B is larger.

Here is a very similar question about 5 and 8 spices:
https://gre.myprepclub.com/forum/contes ... -8340.html

Thank you.

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!

Nelli : You are correct. We have updated the OA to B.
Thank you.

Thank you sir for the clarification BrushMyQuant

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