
In how many ways can Rs.16 be divided into 4 persons when none of them get less than Rs.3?
A.70
B.35
C. 64
D.190
Answer
164.1k+ views
Hint: In this question, first of all, we will determine the number of ways to divide the Rs.16 into 4 persons. So, to find the number of ways to divide Rs.16 into four persons, we will use the concept of permutation and combinations.
Formula Used:
$ \dfrac{{(n + r - 1)!}}{{(r - 1)!n!}}$
Complete step by step Solution:
Let us assume that ${a_1},{a_2},{a_3}\:and \:{a_4}$ are the four persons. Rs.16 is to be divided between them in such a manner that none of them gets less than Rs.3
So, we can say that
\[ \Rightarrow \begin{array}{*{20}{c}}
{{a_1} + {a_2} + {a_3} + {a_4}}& = &{16}
\end{array}\]
And the condition is that none of them get less than Rs.3
Therefore, Rs.3 will be fixed which each person will get. It means that Rs.12 will be divided among them but the remaining Rs.4 will be distributed in such a manner that each person may get Rs.1 or less than Rs.1
Therefore,
Now, consider that \[{a_1}\]get Rs (\[p + 3\]). Similarly,
\[{a_2}\], \[{a_3}\]and \[{a_4}\]will get (\[q + 3\]), (\[r + 3\]), and (\[s + 3\]) respectively.
Now,
\[\begin{array}{*{20}{c}}
{ \Rightarrow (p + 3) + (q + 3) + (r + 3) + (s + 3)}& = &{16}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow p + q + r + s}& = &{16}
\end{array} - 12\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow p + q + r + s}& = &4
\end{array}\]………………(A)
Now, we know that
Number of ways of distribution = $\dfrac{{(n + r - 1)!}}{{(r - 1)!n!}}$
From equation (A),
\[\Rightarrow \begin{array}{*{20}{c}}
n& = &4
\end{array}and\begin{array}{*{20}{c}}
r& = &4
\end{array}\]
Therefore,
Number of ways of distribution = \[\dfrac{{(4 + 4 - 1)!}}{{(4 - 1)!4!}}\]
The number of ways of distribution = \[35\].
So, the final answer will be 35.
Hence, the correct option is (B).
Note: It is important to note that each person will get a fixed Rs.3 and some Rs will be added to distribute Rs.16 between 4 persons. Rs.12 will be divided equally but the remaining Rs.4 may be divided equally or not. That is why we can say that the whole calculation will be done to divide Rs.4 among four persons.
Formula Used:
$ \dfrac{{(n + r - 1)!}}{{(r - 1)!n!}}$
Complete step by step Solution:
Let us assume that ${a_1},{a_2},{a_3}\:and \:{a_4}$ are the four persons. Rs.16 is to be divided between them in such a manner that none of them gets less than Rs.3
So, we can say that
\[ \Rightarrow \begin{array}{*{20}{c}}
{{a_1} + {a_2} + {a_3} + {a_4}}& = &{16}
\end{array}\]
And the condition is that none of them get less than Rs.3
Therefore, Rs.3 will be fixed which each person will get. It means that Rs.12 will be divided among them but the remaining Rs.4 will be distributed in such a manner that each person may get Rs.1 or less than Rs.1
Therefore,
Now, consider that \[{a_1}\]get Rs (\[p + 3\]). Similarly,
\[{a_2}\], \[{a_3}\]and \[{a_4}\]will get (\[q + 3\]), (\[r + 3\]), and (\[s + 3\]) respectively.
Now,
\[\begin{array}{*{20}{c}}
{ \Rightarrow (p + 3) + (q + 3) + (r + 3) + (s + 3)}& = &{16}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow p + q + r + s}& = &{16}
\end{array} - 12\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow p + q + r + s}& = &4
\end{array}\]………………(A)
Now, we know that
Number of ways of distribution = $\dfrac{{(n + r - 1)!}}{{(r - 1)!n!}}$
From equation (A),
\[\Rightarrow \begin{array}{*{20}{c}}
n& = &4
\end{array}and\begin{array}{*{20}{c}}
r& = &4
\end{array}\]
Therefore,
Number of ways of distribution = \[\dfrac{{(4 + 4 - 1)!}}{{(4 - 1)!4!}}\]
The number of ways of distribution = \[35\].
So, the final answer will be 35.
Hence, the correct option is (B).
Note: It is important to note that each person will get a fixed Rs.3 and some Rs will be added to distribute Rs.16 between 4 persons. Rs.12 will be divided equally but the remaining Rs.4 may be divided equally or not. That is why we can say that the whole calculation will be done to divide Rs.4 among four persons.
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