SOLUTION: There are 2 Senators from each of 50 states. We wish to make a 3-Senator committee in which no two members are from the same state. How many ways can the 3-Senator committee be for

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Question 1169916: There are 2 Senators from each of 50 states. We wish to make a 3-Senator committee in which no two members are from the same state. How many ways can the 3-Senator committee be formed such that no two Senators are from the same state?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

1-st senator in the committee can be any one of 100 senators, giving 100 options.


2-nd senator can be any one of 98 senators from other states, giving 98 options.


3-rd senator can be any of 96 senators from remaining 48 states, giving 96 options.


In all,  there exists  100*98*96 = 940800  different ways.    ANSWER

Solved.

================

My note after reading the post by @Math_tutor2020. 

    @Math_tutor2020 correctly pointed that my number must be divided by 3! = 6:


        940800/6 = 156800.


    I agree with it, in full (!)



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

There are three slots or seats to fill on this committee.

For the first slot, we have 100 choices since we have 2*50 = 100 senators.
After that person is picked, there are 98 choices left. Why 98 and not 99? Because we cannot pick a senator from the same state as the first person selected.

For example, let's say we pick a senator from Michigan to fill the first seat. The other senator from Michigan cannot be selected if we want this three-person committee to not have two members from the same state.

So again we have 98 choices for the second slot.

Then whoever goes in the second slot will lower the amount of choices down to 96 for similar reasoning as before.

Overall we have 100*98*96 = 940,800 permutations possible

Since order doesn't matter, we divide by 3! = 3*2*1 = 6 to get (940,800)/6 = 156,800

Note how there are 6 ways to arrange a group of three items
{A,B,C}
{A,C,B}
{B,A,C}
{B,C,A}
{C,A,B}
{C,B,A}
All six sets listed above are the same set since all have A,B,C in some order. The order doesn't matter on a committee. No member outranks another. All that matters is the group overall.

Answer: 156,800
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