Question

100 identical objects are distributed among 10 persons in which ${{1}^{st}}$ person gets at least one object, ${{2}^{nd}}$ person gets at least 2 objects, ${{3}^{rd}}$ person gets at least 3 objects, …., ${{10}^{th}}$ person gets at least 10 objects, then total number of ways that the objects are distributed is
(a) ${}^{55}{{C}_{10}}$
(b) ${}^{55}{{C}_{9}}$
(c) ${}^{54}{{C}_{9}}$
(d) ${}^{54}{{C}_{44}}$

Answer 1

We solve this problem first by giving all the 10 persons at least how many they can get. After that we calculate the number of ways of distributing the remaining identical objects to 10 persons. For finding this number of ways of distributing the remaining identical objects to 10 persons we use a standard result that is if $'n'$ identical objects are distributed among $'r'$ distinct places then the number of ways of such distribution is given as ${}^{n+r-1}{{C}_{r-1}}$.

Complete step-by-step answer:
We are given that there are a total of 100 identical objects.
Let us assume that there are 10 places representing the 10 persons.
We are given that ${{1}^{st}}$ person gets at least one object, ${{2}^{nd}}$ person gets at least 2 objects, ${{3}^{rd}}$ person gets at least 3 objects, …., ${{10}^{th}}$ person gets at least 10 objects.
Now, let us give the 10 places the minimum number of objects they can get that is let us give ${{1}^{st}}$person one object, ${{2}^{nd}}$ person 2 objects, ${{3}^{rd}}$ person 3 objects, …., ${{10}^{th}}$ person 10 objects.
Here, we can say that the above distribution can be done in one way only because all the objects are identical so distributing them in this specific order can be done only in one way.
Now, let us calculate the number of objects remained after distribution of objects as mentioned above.
So, the number of objects distributed in above distribution is given as

& \Rightarrow N=1+2+3+......+10 \\\ & \Rightarrow N=55 \\\ \end{aligned}$$ Here, we can say that 55 objects have been already distributed in one way. Now, the remaining objects is calculated as $$\begin{aligned} & \Rightarrow R=100-N \\\ & \Rightarrow R=100-55 \\\ & \Rightarrow R=45 \\\ \end{aligned}$$ So, we can say that there are 45 objects left to be distributed among 10 places. Let us assume that the number of ways of distributing 45 identical objects among 10 places as $$'T'$$. We know that if $$'n'$$ identical objects are distributed among $$'r'$$ distinct places then the number of ways of such distribution is given as $${}^{n+r-1}{{C}_{r-1}}$$. By using the above formula we get the number of ways of distributing 45 identical objects among 10 places is given as $$\begin{aligned} & \Rightarrow T={}^{45+10-1}{{C}_{10-1}} \\\ & \Rightarrow T={}^{54}{{C}_{9}} \\\ \end{aligned}$$ Therefore, we can say that the number of ways of distributing 45 identical objects among 10 places is $${}^{54}{{C}_{9}}$$. So, option (c) is the correct answer. **So, the correct answer is “Option (c)”.** **Note:** Students may make mistakes in the first part of the number of ways of distributing the minimum number of objects each person can get. While distributing $${{1}^{st}}$$ person one object, $${{2}^{nd}}$$ person 2 objects, $${{3}^{rd}}$$ person 3 objects, …., $${{10}^{th}}$$ person 10 objects students take the number of ways as selecting respective objects from 100 objects that is number of ways of distributing in this way is given as $$\Rightarrow n={}^{100}{{C}_{1}}+{}^{100}{{C}_{2}}+{}^{100}{{C}_{3}}+.......+{}^{100}{{C}_{10}}$$ This gives the wrong answer because the objects are all identical number of ways of distributing $${{1}^{st}}$$person one object, $${{2}^{nd}}$$ person 2 objects, $${{3}^{rd}}$$ person 3 objects, …., $${{10}^{th}}$$ person 10 objects is only ‘1’ way. So, we can say $$n=1$$ is the correct answer. The above mentioned parts need to be taken as the objects are identical.