9 people sit in a row. 2 dressed in Red, 7 blue and 14 yellow. What is the P that at least 2 guys in yellow will sit next to a another in yellow? 1There are $\frac{21!}{2!3!} = 120$ total positions (disregarding order within same colour). I imagine labelling the people Y (yellow) and NY (not yellow), so I imagine I have $4$ copies of the letter Y and $5$ of NY. So I draw out $9$ slots and want to arrange so that at least $2$ Y are together. 
I get $30$ arrangements, if I consider $2$ lots of YY sitting in different positions.
$47$, if only one block of YY and the rest of Y’s are sitting separately.
$30$ if YYY are all sat next to each other.
And $6$ if all of them are sitting next to each other. Total of $113$ arrangements.
Now considering each Y and each NY to be distinct each of the $113$ arrangements allow $5!$ ordering of NY(non yellow) and $4!$ orderings of yellow.
So we get $113 \times 5! \times 4!$ ways of ordering at least one yellow shirted person next to another.
And so the total probability is:
 $$\frac{113 \times 5! \times 4!}{ 1260} = \frac{113}{126}$$
 which seems to me a totally wrong answer.
Any suggestions?
 A: You are thinking about this problem in a very complicated way which will probably lead to missed cases and the wrong answer. I think there is a better way to approach this problem.
There are $1260$ arrangements in all. Therefore, we can find the number of arrangements such that all of the $Y$s are separated by at least one person, which is the opposite of what we want, and then subtract that from $1260$.
To find the number of arrangements where the $Y$s do not sit next to each other, consider the following:
$$Y \ (NY) \ Y \ (NY) \ Y \ (NY) \ Y$$
Here, we have two $NY$s left to place. We can place them outside of the $Y$s or in between them, giving us five places. This means, by stars and bars, there are ${5+2-1 \choose 5-1}={6 \choose 4}$ ways to place the $NY$s. Then, to order the $NY$s, there are $\frac{5!}{2!3!}$ ways to do that, disregarding order with the same color. Thus, this is the number of ways to arrange the $Y$s so they are not adjacent:
$$\frac{5!}{2!3!}\cdot {6 \choose 4}=150$$
Now, this means there are $1260-150$ ways to arrange the $Y$s so they can sit next to each other. Therefore, the probability that this happens is:
$$\frac{1260-150}{1260}=\frac{37}{42}$$
A: Hint...consider when none of the yellows are next to each other, which can happen in $\binom 52 \times \binom 64$ ways...can you finish?
A: Alright, we have 4 times Y and 5 times N. 
There are $9\choose 4$ arrangements in total, including those with YY somewhere
If we replace each occurrance of N followed by Y with an X, then a non-favourable outcome is 


*

*either Y followed by any combination of three X and two N; there are $5\choose 3$ of these

*or any combination of for X and one N; there are $5\choose 4$ of these


In summary, there are $15$ non-favourable out of $126$ in total, hence the probability is $1-\frac 5{42}$.
A: As others have indicated, it is easier to count the number of cases in which no two people in yellow shirts are adjacent.
We focus only on the shirts, not who is wearing them, which reduces the problem to arranging red, blue, and yellow shirts.
For the denominator, we arrange the shirts in a row.  We choose two of the nine positions for the red shirts, three of the remaining seven positions for the blue shirts, and the four remaining positions for the yellow shirts, giving a total of 
$$\binom{9}{2}\binom{7}{3}\binom{4}{4}$$
possible arrangements.
Next, we count arrangements in which no two yellow shirts are consecutive.
First, we line up the red and blue shirts in a row.  We choose two of the five positions for the red shirts and the remaining three positions for the blue shirts.  There are $\binom{5}{2}\binom{3}{3}$ such arrangements.
For a given arrangement of red and blue shirts, we now have six spaces in which to place the yellow shirts, four between successive shirts and the two spaces at the ends of the row.  To ensure that no two yellow shirts are placed in adjacent positions, we choose four of these six spaces in which to place one yellow shirt each, which can be done in $\binom{6}{4}$ ways.
Hence, the number of arrangements in which no two yellow shirts are consecutive is 
$$\binom{5}{2}\binom{3}{3}\binom{6}{4}$$
Thus, the probability that no two yellow shirts are consecutive is 
$$\frac{\dbinom{5}{2}\dbinom{3}{3}\dbinom{6}{4}}{\dbinom{9}{2}\dbinom{7}{4}\dbinom{3}{3}}$$
We want the complementary probability that at least two yellow shirts are consecutive, which is 
$$1 - \frac{\dbinom{5}{2}\dbinom{3}{3}\dbinom{6}{4}}{\dbinom{9}{2}\dbinom{7}{4}\dbinom{3}{3}}$$
A: There are $\binom{5+4}4$ sums $a+b+c+d+e=5$ where the $a,b,c,d,e$ are nonnegative integers. Here $a$ stands for the number of non-yellows on the left of the utmost left yellow, $b$ for the number of non-yellows between the utmost left yellow and the yellow closest to the utmost left yellow, et cetera. 
There are $\binom{2+4}{4}$ sums $a+b+c+d+e=5$  where the $a,e$ are nonnegative integers and $b,c,d$ are positive integers. This because it equals the number of sums $a+b'+c'+d'+c=2$ where the $a,b',c',d',e$ are nonnegative integers.
That gives probability $$1-\frac{\binom64}{\binom94}=
\frac{37}{42}$$
A: To count unfavorable ways, arrange the reds and blues in $\dfrac{5!}{3!2!} = 10$ ways,
and insert the yellows at the uparrows in $\binom64 = 15$ ways
$\uparrow\fbox{red or blue}\uparrow\fbox{red or blue}\uparrow\fbox{red or blue}\uparrow\fbox{red or blue}\uparrow\fbox{red or blue}\uparrow$
$Pr = 1 - \dfrac{10*15}{1260} = \dfrac{37}{42}$
A: As pointed out in the other answers, it is easier to solve this problem by finding the probability that no two people wearing yellow are sitting together and then subtracting this from 1.

Using your approach, though, with 4 Y's and 5 N's there are $\dbinom{9}{4}$ total arrangements, 
and we can count the arrangements where two Y's are together by considering your 4 cases:
1) If we have YY and YY, we need an N between them, and there are $\dbinom{6}{2}$ ways to arrange 
$\hspace{.2 in}$the remaining 4 N's in the 3 gaps.
2) If we have YY, Y, Y, there are 3 ways to order these elements; and if we insert N's between them, 
$\hspace{.2 in}$then there are $\dbinom{6}{3}$ ways to arrange the remaining 3 N's in the 4 gaps.
3) If we have YYY, Y, there are 2 ways to order them; and after inserting an N between them, 
$\hspace{.2 in}$there are $\dbinom{6}{2}$ ways to arrange the remaining 4 N's in the 3 gaps.
4) If we have YYYY, then there are 6 possibilities.
Therefore the probability of having two Y's together is equal to
$\displaystyle\frac{\binom{6}{2}+3\binom{6}{3}+2\binom{6}{2}+\binom{6}{1}}{\binom{9}{4}}=\frac{111}{126}=\frac{37}{42}$
