# Multinomial distribution to Binomial distribution?

We have a multinomial distribution on $X=(X_1,\ldots,X_r)$ with the parameters $(n,p_1,\ldots,p_r)$

Now we will have to determine the distribution of $X=(X_1+X_2,X_3,\ldots,X_r)$

Does that mean we get from the multinomial distribution to the binomial distribution?

The only difference I notice that the first two parameters of the second tuple are added together. $X_1+X_2$

If $X=(X_1,\ldots,X_r)$ has a multinomial distribution, then each of the components $X_1,\ldots,X_r$ has a binomial distribution.
You're distributing $n$ objects into $r$ bins. For each object, the probability that it falls into the $k$th bin is $p_k,$ for $k=1,\ldots,r.$ The number of objects that fall into the $k$th bin is $X_k,$ for $k=1,\ldots,r.$ So $X_1+X_2$ is the number of objects falling into either of the first two bins. The probability that an object falls into either of the first two bins is the sum of the probabilities of its falling into those two bins, i.e. it is $p_1+p_2.$ In effect, you've simply joined those two bins together, so now you have $r-1$ bins, with probabilities $p_1+p_1,p_3,p_4,\ldots,p_r$ of an object falling into them. Therefore the distribution of $(X_1+X_2,X_3,X_4,\ldots,X_r)$ is multinomial with parameters $(n,p_1+p_2, p_3, p_4, \ldots, p_r).$ And as before, each component separately has a binomial distribution.
It means that you step over from $r$-nomial distribution to $r-1$-nomial distribution.
The parameters are now $(n,p_1+p_2,p_3,\dots,p_r)$.