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First lets define our n choose k notation.

$${n \choose k} = \frac{n!}{k! (n-k)!}$$

Next lets setup our answer which we have as "99 choose 6 multiplied by 99 choose 1"

$${99 \choose 6} \cdot {99 \choose 1}$$

Now lets expand that to our above equation

$$\frac{99!}{6! (99-6)!} \cdot \frac{99!}{1! (99-1)!}$$

Now lets simplify the subtraction.

$$\frac{99!}{6! \cdot 93!} \cdot \frac{99!}{1! \cdot 98!}$$

We know the factorial of 1 is just one based on the idea that a factorial are all the numbers up to itself multipled. There we can simplify further to

$$\frac{99!}{6! \cdot 93!} \cdot \frac{99!}{1 \cdot 98!}$$

$$\frac{99!}{6! \cdot 93!} \cdot \frac{99!}{98!}$$

So this is probably where youll get stuck, but think about what a factorial means..

$$99! = 99 \cdot 98 \cdot 97 \cdot 96 \cdot\ ...\ 2 \cdot 1$$

So now if we imagine the larger numbers expanded something should become obvious if you remember your fractions

$$\frac{99!}{6! \cdot 93!} \cdot \frac{99 \cdot 98 \cdot 97 \cdot 96\ ...}{98 \cdot 97 \cdot 96\ ...}$$

Notice how the top part of the fraction is pretty much multiplying all the same numbers as the bottom part except it just has that extra 99 multiplied at the end because the factorial was one larger. Remember from your fractions that when you have common factors on the top and the bottom of a fraction they cancel eachother out.. thats why 4/4 is the same as 1/1, the fours cancel out and leave ones. so the 98 in the top and bottom cancel as do the 97, etc all the way to one. so the right hand fraction now becomes just 99

$$\frac{99!}{6! \cdot 93!} \cdot 99$$

If we do the same on the left hand fraction then all the factors from 93 to one on the bottom would cancel out with the ones on the top, so we would be left with...

$$\frac{99 \cdot 98 \cdot 97 \cdot 96 \cdot 95 \cdot 94}{6!} \cdot 99$$

Likewise expand the $6!$

$$\frac{99 \cdot 98 \cdot 97 \cdot 96 \cdot 95 \cdot 94}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot 99$$

from here on out its just simple math.

$$\frac{806781064320}{720} \cdot 99$$

$$110932396344$$

So there are 110,932,396,344 different combinations.

Tue, 01 Dec 2020 02:19 GMT