Number of ways to arrange stripes on a flag I came across this permutations and combinations problem.
A Flag has 4 stripes. We can fill those 4 stripes with 7 colours: VIBGYOR. In how many ways can the colours be filled such that two of the colours on the flag are compulsorily V and B?
This is how I approached the problem.
Out of the 4 stripes, 2 are already filled by V and B, so only the other 2 have to be filled. Let each stripe be numbered 1, 2, 3, 4. Assuming that we can't repeat colours, I can put 5 x 4 colours in the two stripes which are empty.
Now to fill the V and B. There can be 6 different cases.
V and B in 1 and 2 respectively
V and B in 2 and 1 respectively
V and B in 2 and 3 respectively
V and B in 3 and 2 respectively
V and B in 4 and 3 respectively
V and B in 3 and 4 respectively
In each of these 6 cases, I can arrange the other two colours in 5 x 4 ways. Thus the total number of ways the colours can be arranged according to the question will be 5 x 4 x 6, which is 120.
The answer given, for some reason, is 240.
Could anyone please point out what I'm missing in my approach above and/or suggest a better one?
Thanks!
 A: Well done, the only problem is that you are placing $V$ and $B$ together (is not assumed in the problem). In other terms, just pick the two colors from the remaining $5$ options in $\binom{5}{2}=10$ ways and permute the $4$ colors in $4!=24$ ways.
A: I think that there is someting missing in the question . It should have been stated whether repetition is allowed or not . If the repetition is not allowed  ,then @Phicar 's answer is elegant !
However , if repetition is allowed , then answer will be different. I am posting this answer , because the question is not clear to show its restriction and you may want to see what happens when the repetition allowed.
First of all , the most trustable method for this question is exponential generating functions. It is said that the colors $V$ and $B$ must be in the strips of flag. So , their exponential form will be $$\bigg( x + \frac{x^2}{2} + \frac{x^3}{6} \bigg)$$ The expoenetial form of the rest will be $$\bigg(1+ x + \frac{x^2}{2} + \bigg)$$ Now  , find the coefficient of $x^4$ in the expansion of $$\bigg( x + \frac{x^2}{2} + \frac{x^3}{6} \bigg)^2 \bigg(1+ x + \frac{x^2}{2}  \bigg)^5$$ and  multiply it by $4!$ such that https://www.wolframalpha.com/input/?i=expanded+form+of+%281%2Bx+%2B+x%5E2+%2F+2+%29%5E5++%28x+%2B+x%5E2+%2F+2+%2B+x%5E3+%2F+6+%29%5E2+
$$\frac{217}{12} \times 4! = 434$$
Or, you can use principle of inclusion-exclusion such that arrange all strips when repetiton allowed by $7^4$ ways . However , we want that $V$ and $B$ must exist , so we should subtract the situations where $V$ and $B$ do not exists from $7^4$.
We can do it by all arrangements where $V$ do not appear + all arrangements where $B$ do not appear  - all arrangements where $V$ and $B$ do not appear .It is equal to $6^4 + 6^4 - 5^4$
Then , $$7^4 - (6^4 + 6^4 - 5^4) =434$$
A: 
Could anyone please point out what I'm missing in my approach above and/or suggest a better one?

Talking about your approach exactly, you are missing 2 in the 3rd paragraph.
V and B in 1 and 4 respectively and V and B in 4 and 1 respectively.
