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Re: Contestants at a baking contest must use between 5 and 8 of
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03 Feb 2023, 22:14
Given that the contestants at a baking contest must use between 5 and 8 of 10 possible ingredients.
Now the number of ingredients which we can choose can be between 5 and 8(assuming inclusive) => 5, 6, 7, 8
If we are choosing 5, then we can choose 5 out of 10 on 10C5 ways = 10!(5!)∗(10−5)! = 10!5!∗5! = 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 = 10!(5!)∗(10−6)! = 10!6!∗4! = 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 = 10!(7!)∗(10−7)! = 10!7!∗3! = 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 = 10!(8!)∗(10−8)! = 10!8!∗2! = 10∗9∗8!8!∗1∗2 = 45 ways
Quantity A: The number of ingredients that must be used to get the smallest possible number of different combinations
Smallest number is produced when we take number of ingredients as 8
=> Quantity A = 8
Quantity B: The number of ingredients that must be used to get the largest possible number of different combinations.
Greatest number is produced when we take number of ingredients as 5
=> Quantity B = 5
Clearly, Quantity A > Quantity B
So, Answer will be A
Hope it helps!