jfan17

Start with a "reference bead" that we can use to indicate the position of each bead, with the reference bead being one of the distinct colored beads. Next, realize that of the 5 remaining beads left, the possible ways to select the same-colored beads is \({5 \choose 2}\)

= 10 ways to pick the pair. After you pick where the pair goes, the remaining beads have 4! possibilities to go to(do you see why? The first bead has 4 places, the second has 3, etc.) The total number of possibilities is then 10 * 24 = 240 arrangements. If you see something wrong with my solution, plz lmk. I haven't been the best historically with counting  ;).

ok I'm not really sure why, but web2.0 calc keeps constantly flagging my responses for this question for moderation. I'll just explain the basic jist of it. This problem employs a common counting method known as "stars and bars." with stars and bars, you can find out how many ways there are to distribute n things into k groups. For stars and bars, there are two forms of it: non negative and positive stars and bars(in other words, non negative is where a group is allowed to have 0 things in it, and positive at least 1 thing). This problem is asking for the non negative form of stars and bars, which is :

\({n+k-1 \choose n}\) With this formula, the problem becomes a lot easier!

This is a really common method of counting known as "stars and bars". If you want more reference information, I'd highly recommend looking to wikipedia or other sources online. The basic idea is that we can think of 50 votes as "50 balls" or "stars" laid out in a row.

These 50 stars are divided by 3 "bars", which form 3 groups. You can picture it like this:

if the O's are balls, and the |'s are dividers, we have:

O O O O O O | O O O O O | O O O O | O O .. ..... x 50.

These bars divide how many votes each of the four candidate gets(can you see why there are 3 bars? because if n is the number of bars there are, n+1 regions are formed from n bars). 

Because we want to see how many votes each candidate could get, the answer is stars and bars' non negative form(there are two forms of stars and bars, watch out!)

With stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

This simplifies to:

53*26*17 = 23426 totals

1.  This is a textbook case of a method of counting known colloquially as "stars and bars" or "balls and urns"(I'd recommend you to search it up online). We can think of the 50 votes as 50 balls laid out in a row, divided by 3 "dividers" which separate the balls into 4 separate regions for each candidate, with each region being how many votes a candidate gets. Using stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

This simplifies to:

53*26*17 = 23426 totals

For stars and bars:

like the above example, just think of the votes as 50 "balls" that are divided by 3 dividers, which create 4 regions for vote allocation(for 4 candidates). the formula for stars and bars(non negative, meaning 0 is allowed) is:

(n + k -1) choose (n) 

where n is the number of things you want to distribute or "stars", and k is the number of groups you want to distribute those stars into.

I used a method of counting known as "stars and bars". I'd recommend you search it up! It's really interesting imo

https://web2.0calc.com/questions/can-t-solve-this-blame-skyrim 

This question has been posted maybe 6 times already in the past day - 2. I solved part of it here

That wouldn't really make sense tho, bc how would the forum know if someone was sure or not? Just food for thought.

so here's the link to my solution for part B:

https://web2.0calc.com/questions/can-t-solve-this-blame-skyrim

I don't really know why, but the forum keeps flagging my solution to part a for moderation(even when I post it in snippets) so I'll have to wait on that. In the meantime, you could look at how I solved part B(which in my opinion is harder than part A!). Using that, try solving Part A yourself!

1.  This is a textbook case of a method of counting known colloquially as "stars and bars" or "balls and urns"(I'd recommend you to search it up online). We can think of the 50 votes as 50 balls or stars laid out in a row, divided by 3 "dividers" or bars which separate the balls into 4 separate regions for each candidate, with each region being how many votes a candidate gets. With this, we can think of how many possibles "spaces" you can choose for the dividers to go in. Using stars and bars, we get the equation:

(50+4-1) c (50),  or (53c50). This gives us 53*52*51/3!

This simplifies to:

53*26*17 = 23426 totals

2. So as for Part b, it becomes a little trickier. Imagine the problem as the same as the first part, except there are 4 dividers. You might be wondering why there are four dividers, since that produces 5 "regions", but there are only 4 candidates! However, you need the 5th "region" to count for people who abstain(which is a clever trick you can use for problems like this) from voting for a candidate; you could almost think of the people who abstain as people who vote for another candidate. With this knowledge in hand, the problem becomes a lot like the first problem, and with stars and bars, we get:

(50+5-1) c (50) or (54c50). This gives us 54*53*52*51/4!, or 316251 totals

EDIT: Since this post finally got through, my answer to part A was flagged for moderation, but using stars and bars, we can find the answer to both the first and second parts. However, Part B just puts a "spin" on the stars and bars technique as I mention above