You want to arrange $5$ identical white balls, $4$ identical red balls and $3$ identical blue balls in a row.
a) In how many different ways can this be done?
b) What if we are required that all white balls are placed to the left of any red ball?
c) What if we are required not to have a white and a red ball sit next to each other? (Hint: think of blue balls as separators/slashes)
d) What if no two white balls may sit next to each other?
e) What if every two blue balls are separated by at least three non-blue balls?
Here is what I have done so far:
a) I took the total number of balls with a factorial divided by the number of balls per each type each with a factorial: $\dfrac{11!}{5!4!3!}$
b) I'm not really sure how to do this…I was going to make the White balls and the Red balls a block of balls in which the whites precede the reds but I'm not so sure how to develop the multiple cases in which the blues are in between the whites and the reds…etc.
c) I attempted to treat the blues as a divider between the whites and the reds and then ultimately I treated the reds and the blues as a divider of the whites. So I had: Red Blue White Blue Red Blue White where whites can be placed where it says white and reds and blues can be placed where it says red and blue. This way blues serve as a divider and reds and whites are never consecutive. Thus I could place five whites in two spots and would have: ${5+2\choose 2}$ = ${7\choose 2}$ and the reds I would have: ${4+2\choose 2}$ = ${6\choose 2}$. I'm wondering if then I should multiply these two choose functions or if this is correct at all…
d) I took the total number of cases and subtracted the number of bad cases to get the number of desired cases:
Total number of cases: $\dfrac{12!}{5!4!3!}$
Put two Ws next to each other: $\dfrac{11!}{3!4!3!}$
Put three Ws next to each other: $\dfrac{10!}{2!4!3!}$
Put four Ws next to each other: $\dfrac{9!}{4!3!}$
Put five Ws next to each other: $\dfrac{8!}{4!3!}$
So we would have: $\dfrac{12!}{5!4!3!}$ – $\dfrac{11!}{3!4!3!}$ – $\dfrac{10!}{2!4!3!}$ – $\dfrac{9!}{4!3!}$ – $\dfrac{8!}{4!3!}$ ways
e) I started by separating the blue balls by three other balls however I'm not sure because it says only three balls at the least separate the blue balls so I'm unsure how to go about including more than three balls separating the blue balls…
Best Answer
Just answering part b, but this is too long for a comment.
For “b) What if we are required that all white balls are placed to the left of any red ball?”:
You are right to consider the white balls (there are $5$) and the red balls (there are $4$) together. Call them pink. Given an arrangement of $9$ pink balls and $3$ blue balls, you can make the first $5$ pink balls white and the last $4$ pink balls red to obtain an arrangement of the balls where the white balls all come before the red balls. Every arrangement of the red, white, and blue balls with all white balls left of any red ball can be obtained uniquely in this way, so there is a one-to-one correspondence between the “whites before reds” arrangements and the “pinks and blues” arrangements. Thus the number you seek is the same as the number of arrangements of $9$ pink and $3$ blue balls.