How many possible sets of 25 can I get from 200?

first let's solve a mini-problem.

how many ways can you divide ABCD into 2 distinct parts ?

let us start with 4c2*2c2 = 4!/(2!)^2 = 6, the way the first answerer has done.

unfortunately, this is NOT correct because it contains permutations of the 2 parts, eg

AB|CD & CD|AB will be counted separately.

to correct for this, we need to further divide by 2! to get 3 as the ans

the 3 unique partitions are AB|CD, AC|BD & AD|BC

so on the larger scale of 200 to be partitioned into 8 groups, the formula will be

200c25*175c25*150c25* ........ *25c25 / 8!

= 200!/((25!)^8*8!)

≈ 5.837 *10^168

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i know this problem seems hard but you just have to simply it. you need 4 groups because 100/25 = 4. how many ways can you arrange these groups. remember that number!!! now within each group you need to arrangehave 25 balls of that you can only choose 8.. this equation can help N! / ( (N - h)! * h! ).............the ! point means factorial. factorial means that the number is really the product of all its intergers before it up to one example 4! = 4*3*2*1 = 24 3! = 3*2*1 = 6.......in the equation N stands for the number of posible choices for your problem it is 25. h stands for the spots available for you its 8. plug in the number and you should get how many ways those balls can be arranged within the group.........Lastly multiply the number i told you to remember times the number that you calculated and that should give you your final answer

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C(200,25) * C(175,25) * C(150,25) * C(125,25) * C(100,25) * C(75,25) * C(50,25) * C(25,25)

= 200! / ((25!)^8)

=~ 2.35 * 10^173