A bag contains 5 red, 6 blue, and 4 yellow marbles

Question

A. How many ways can marbles be drawn if at least 1 must be drawn?

What I’ve done to attempt to solve this question A.

Just imagine that the marbles aren’t actually marbles, but rather switches. The first 5 switches are representative of “red marbles” and if all 5 of them are switched on, then you have taken all your red marbles. The next 6 are marked blue, and so forth therefore the remaining 4 are yellow. How many ways can the switches be setup? 2^15, so therefore 32,768. However the question asks that you must select at least one marble, so you can’t switch all the switches to off, and so there would be one less, 32,767 ways

However, I am not sure if my approach is correct because due to the wording of the question I believe that order does not matter and therefore my analogy is incorrect.

I don’t believe it matters if you drew 2 marbles first or last or as the 6th and 7th, as 2 marbles would still be 2 marbles. Therefore, the answer might be a lot less than 32,767.

How do I proceed?

The second question is:

B.) How many ways can 5 marbles be drawn if at least 2 are the same colour? With this question I am completely stuck.

The third and final question I am asking as more of a check rather than help.

C.) How many ways can 5 marbles be drawn out at random of at least one of them were yellow? My answer: (15 Choose 5) - (11 Choose 5) = 15!/(5!10!) - 11!/(5!6!) = 1/5![(1514131211) - (1110987)] = (11/5!)[(15141312 - 10987)] = 11/120[ 32760 - 5040] = 2772011/120 = 277211/12 = 23111 = 2541 ways

I’d appreciate the help!

Answer 1

I am assuming that the order of drawing is not important. IE We just care about which marbles we have drawn.
If order is important, please state that.

Hint for A): How many red / blue / yellow marbles did you draw?

$ 6 \times 7 \times 5 - 1 $

Hint for B): If you draw 5 marbles, then at least 2 of them must be the same color. Thus the additional condition is a red herring.

Note that we cannot draw 5 yellow marbles.

$ {7 \choose 2} - 1$

Hint for C): This is a standard Stars and Bars question.

{ 6 \choose 2 }

Answer 2

I agree with Calvin Lin on part A - the only decision here on an unordered choice is how many of each colour we draw, and $5\times 6\times 4$ allows the drawing of zero marbles, so we just need to exclude that one option, $5\times 6\times 4-1$.

For part B, we need to think about the different combinations that are possible for $5$ marbles. I don't understand why the "at least $2$ of the same colour" condition is present, since this is inevitably fulfilled. We can use a sticks & stones approach and just remove the ineligible option of $5$ yellow marbles. For variety I'll build it up from the different multiplicity options - we can have $5x$, $4x+y$, $3x+2y$, $3x+y+z$ and $2x+2y+z$ marbles for some combination of marble colours $x,y,z$. We can choose these the following ways:

• $5x: 2$ options (not yellow)
• $4x+y: 3\times 2 = 6$ options
• $3x+2y: 3\times 2 = 6$ options
• $3x+y+z: 3$ options (choose $x$)
• $2x+2y+z: 3$ options (choose $z$)

giving a total of $20$ options, agreeing with the $\binom 72-1$ version. The detail breakdown approach above is useful if we also have to think about order.

Picking the relevant parts of this detail to answer part C, the $4x+y$ and $3x+2y$ options reduce to $4$ options (pick which slot is yellow, choose the other colour) and $5x$ vanishes, giving $14$ options.