Factorials approximation problem

AI Thread Summary
The discussion centers on simplifying the expression [(N+Q)!Q!]/[(Q+1)!(N+Q-1)!] to show its equivalence to (N+Q)/(Q+1) when N and Q are large. Participants suggest breaking the expression into two fractions: Q!/(Q+1)! and (N+Q)!/(N+Q-1)!. The first fraction simplifies by recognizing that (Q+1)! is Q! multiplied by (Q+1), while the second fraction simplifies to (N+Q) when considering factorial properties. The cancellation of terms leads to the desired result, demonstrating the approximation for large values of N and Q. This analysis confirms the validity of the lecturer's notes.
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Homework Statement


How is,

[(N+Q)!Q!]/[(Q+1)!(N+Q-1)!] equal to (N+Q)/(Q+1) when N,Q>>1 ??

It looks like the Q!/(N+Q-1)! cancels but i don't see how, I am going from my lecturers notes here.

Homework Equations





The Attempt at a Solution

 
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Break it up - look at the fraction as

\frac{Q!}{(Q+1)!} \cdot \frac{(N+Q)!}{(N+Q-1)!}

For the first fraction, ask yourself "what number you would multiply Q! by to get (Q+1)!"?

Do a similar analysis for the second fraction.
 
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