1. For the First Senator, we have 100 ways to choose them.
For the Second Senator, we have 98 senators to choose from(Since we already have 1 senator and we don't want to choose the senator from the same state as the first)
For the Third Senator, we have 96 Senators to choose from(This is using the same logic.)
So, we have 100*98*96 Senators, but we want to know the amount without ordering, so we divide it by 6. This equates to 156800
2. For this problem, we can find all the ways You can form a committee of 3 senators and subtract the ways in which a party holds all 3 seats.
The total amount of ways You can form a committee of 3 senators is 100 choose 3 which is 161700.
The amount of ways to choose 3 Democrats from 53 Democrats is 53 Choose 3, which is 23426
The Amount of ways to choose 3 Republicans from 47 Republicans is 47 Choose 3, which is 16215
Therefore, the amount of ways is 161700-23426-16215 = 122059
We can categorize the different numbers into groups using a counting method called Caseworks.
The thousands digit is set to be 5, since there are no other numbers between 5000-6000 who have thousands digits that aren't 5.
The first category is adding 1's and 2's to make 5.
5122, 5212, and 5221 are all items in the first category.
The second category is adding 1's and 3's
5131, 5113, and 5311 are items in the second category.
The third category is adding 1's and 4's.
5140, 5104, 5014, 5041, 5401, and 5410 are all items in this category.
The 4th category is adding 2's and 3's.
5230, 5203, 5302, 5320, 5023, and 5032 are all items in the 4th category. (Do you sense a pattern regarding Combinatorics?)
There is a pattern regarding the different categories that allows you to finish this without counting everything, though we already have counted everything. Try to find it!
Now we can look at all the numbers we counted and count them. 3 + 3 + 6 + 6 = 18, so there are 18 of them.
Well, a factorial (n!) is a type of equation in which you multiply a number by itself, the number below it, and so on until you reach one. (Ex: 2! = 2 * 1, and 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) For n! = 110, there is no number n, for the closest n is 5, and 5! = 120. Are you sure it's 110 and not 120?