Number of terms in expansion of \(\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{r}}\right)^{\mathrm{n}}\) is (a) \({ }^{(\mathrm{n}+1)} \mathrm{C}_{4}\) (b) \({ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{(\mathrm{r}-1)}\) (c) \(^{(\mathrm{n}-\mathrm{r}+1)} \mathrm{C}_{(\mathrm{r}-1)}\) (d) \({ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}}\)

The multinomial theorem helps us expand expressions of the form \((x_{1}+x_{2}+\ldots+x_{r})^{n}\), where \(r\) is the number of variables and \(n\) is the exponent. The coefficients in the expansion will be in terms of combinations.

The combination formula, denoted as \(\ ^{n}\mathrm{C}_{r}\), is a way of finding the number of different ways to choose r elements from a set of n elements. The formula for a combination is given by: \[ \ ^{n}\mathrm{C}_{r} = \frac{n!}{r!(n-r)!} \] Where \(n! = n(n-1)(n-2) \ldots (1)\) is called the factorial of n.

To find the number of terms in the expansion, we need to find the sum of all possible ways of choosing 'r' elements from the expression. In other words, the combination formula will be applied with a changing r. Since the exponent in our expression is 'n', we understand that the sum of powers of all the variables in each term must equal \(n\). Thus, considering the general term of the expression in the form: \({a}_{1}^{{b}_{1}}{a}_{2}^{{b}_{2}} \ldots {a}_{r}^{{b}_{r}}\) The sum of all possible powers b should equal n: \({b}_{1}+{b}_{2}+\ldots+{b}_{r}=n\)

We can now use the stars and bars method to count the number of different combinations. The stars and bars method is a combinatorial method used to count the number of ways \(n\) things can be distributed into \(r\) distinct categories. In our current problem, we can consider \(n\) stars and \((r - 1)\) bars, which will give us a total of \((n + r - 1)\) elements. In order to find the number of ways to distribute these elements, we need to choose \((r - 1)\) of these elements as bars, and the rest would automatically be stars. The total number of combinations is therefore given by: \[ \ ^{(n+r-1)}\mathrm{C}_{(r-1)} \] Writing all the given choices below: a) \({ }^{(\mathrm{n}+1)} \mathrm{C}_{4}\) b) \({ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{(\mathrm{r}-1)}\) c) \(^{(\mathrm{n}-\mathrm{r}+1)} \mathrm{C}_{(\mathrm{r}-1)}\) d) \({ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}}\)

Comparing our result with the given options, we can see that the correct answer is: (b) \({ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{(\mathrm{r}-1)}\)