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If, of p + q + r things, p are alike and q be alike and the rest different, the total number of combination is
a. $ \left( p+1 \right)\left( q+1 \right){{r}^{2}}-1 $
b. $ \left( p+1 \right)\left( q+1 \right){{r}^{r}}-1 $
c. $ \left( p+1 \right)\left( q+1 \right){{2}^{r}}+1 $
d. $ \left( p+1 \right)\left( q+1 \right){{2}^{r}}-1 $

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Hint: In order to find the solution of this question, we will first find the number of combinations for each group and then we will combine them all to get the correct answer. And we will consider a few cases such as not selecting any of the p,q and r and then generalize the situation. Therefore we have to subtract 1 from the final generalized equation to get the desired result.

Complete step-by-step answer:
In this question, we have been asked to find the number of combinations for p + q + r things, when p are alike, q are alike and the rest all are different. Now, we know that whenever we have to choose from items, we randomly choose one irrespective of the number of ways to pick. And if we have to pick from different items, then we apply combinations, that is $ ^{n}{{C}_{r}} $ .
Now, we will calculate the number of ways of choosing items from p alike items. So, we know that we can either choose 0 or 1 or 2 or more upto p from p items and in each case, there is only 1 possible case of choosing. So, we can say that the total number of ways of choosing from p alike items is (p + 1) ways.
Similarly, we will calculate the number of ways of choosing items from q alike items. So, we know that we can either choose 0 or 1 or 2 or more upto q from q items and in each case, there is only 1 possible case of choosing. So, we can say that the total number of ways of choosing from q alike items is (q + 1) ways.
Now, we have to find the number of ways of choosing items from r different. Now, we know that if we choose 0 items from r, then there are $ ^{r}{{C}_{0}} $ ways to choose and if we choose 1 out of r, then there are $ ^{r}{{C}_{1}} $ ways to choose and if we choose 2 out of r, then there are, $ ^{r}{{C}_{2}} $ ways to choose and more upto r items to choose out of r items, it is $ ^{r}{{C}_{r}} $ ways. So, we can say that, in total there will be $ ^{r}{{C}_{0}}{{+}^{r}}{{C}_{1}}{{+}^{r}}{{C}_{2}}+.........{{+}^{r}}{{C}_{r}} $ ways. Now, we know that $ ^{n}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+.........{{+}^{n}}{{C}_{n}}={{2}^{n}} $ . So, we can say that the total combinations of choosing items from r different items are $ {{2}^{r}} $ ways.
Now, we have been asked to calculate the total number of combinations for p + q + r items. So, we will combine 3 of the situations. So, we get, $ \left( p+1 \right)\left( q+1 \right){{2}^{r}} $ ways but this also includes a case in which none of the 3 is chosen which should not be there, so we have to subtract the total number of ways with 1. So, we get,
Total number of combinations = $ \left( p+1 \right)\left( q+1 \right){{2}^{r}}-1 $
Hence, we can say that option (d) is the correct answer.

Note: While solving this question, we need to remember that whenever we choose 0, 1, 2 or more things from n identical things, then there is one possible way for each case. So, in total it will have only (n + 1) ways to choose items out of n. Also, we need to remember that $ ^{n}{{C}_{0}}{{+}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+.........{{+}^{n}}{{C}_{n}}={{2}^{n}} $ . Also, one can choose option (c) as the correct answer in a hurry as it looks similar to the actual answer, but it is wrong, so one has to choose the option very carefully.
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