I want to understand what approach we would use to figure this out. I didn’t calculate this number instead I used the logic that if the numerator is same then numbers can only be affected by the denominator and bigger the number, bigger its factorial value. So, I just multiplied 5\times316 that gave me 1580 and 7\times314 that gave me 2198. So, denominator of 2nd number is bigger that means its factorial will also be bigger hence overall value of second would be smaller keeping numerator constant but the answer shows quantity B is bigger. How is that so?
Your approach is really smart actually.
Is the denominator just 5 \times 316 or is it 5! \times 316!? they are not quite the same right? The exclamanation marks matter!
I think the intended way of solving the question would be to simplify the fractions on both sides.
For example:
321! = 321 \times 320 \times 319 \times 318 \times 317 \times 316 \times 315...
This can be re-written as 321! = 321 \times 320 \times 319 \times 318 \times 317 \times 316!...
See if you can use that to simplify the fractions and make the comparison easier.
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Actually Quantity A is 321C5 and Quantity B is 321C7. That makes the problem a one-liner.
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Yeah, I was looking for a more time saving method but ig it needs to be solved like this only. Thank you so much!
Yeah you are right. So, could we conclude that if we choose higher number of combinations like as r increases the value of combinations you get also increases? Is this always true?
Up to the half-way point. Try with a small value of n in nCr and see what you get when you change r.
Wow, nice!