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## Main Question or Discussion Point

To calculate the multiplicities of 600 heads in 1000 coin tosses you start with 1000 choose 600 or

1000! / (600! * (1000-600)!) which equals 1000! / (600!) * (400!).

Since you can't calculate this easily, apply Stirling's approx.

N! = N^N * e^(-N) * sqrt(2piN). Applying this to numerator and denominator and leaving out sqrt terms since they are not the problem:

1000^1000 * e^(-1000)

--------------------------------

600^600*e-600 * 400^400 *e-400

Now the exponential terms cancel, but you still can't calculate the remaining terms.

Question: What is the trick to reduce this to something I can calculate? Here's my best attempt.

I tried factoring the bases and canceling terms but that gives

10*10 ^1000 * 10^1000 10^1000

--------------------------------------- = --------------------

10*10 ^600 * 10*10 ^400 * 6^400 * 4^400 6^600 * 4 ^400

You can cancel common factor of 2^1000 the same way giving

5^1000 / (3^600 * 2^400)

I still can't calculate this. Is there another approach or am I missing some other opportunity for canceling?

1000! / (600! * (1000-600)!) which equals 1000! / (600!) * (400!).

Since you can't calculate this easily, apply Stirling's approx.

N! = N^N * e^(-N) * sqrt(2piN). Applying this to numerator and denominator and leaving out sqrt terms since they are not the problem:

1000^1000 * e^(-1000)

--------------------------------

600^600*e-600 * 400^400 *e-400

Now the exponential terms cancel, but you still can't calculate the remaining terms.

Question: What is the trick to reduce this to something I can calculate? Here's my best attempt.

I tried factoring the bases and canceling terms but that gives

10*10 ^1000 * 10^1000 10^1000

--------------------------------------- = --------------------

10*10 ^600 * 10*10 ^400 * 6^400 * 4^400 6^600 * 4 ^400

You can cancel common factor of 2^1000 the same way giving

5^1000 / (3^600 * 2^400)

I still can't calculate this. Is there another approach or am I missing some other opportunity for canceling?