
In how many ways 10 balls can be divided between two boys, one receiving two and other receiving eight balls ?
Answer
521.7k+ views
Hint: As it is obvious that all balls are identical, it does not depend on which order the ball is distributed. So, we had to only assume two boys A and B. And then we had to find a number of ways in two cases ( A - 2 balls, B - 8 balls and A - 8 balls, B - 2 balls).
Complete step-by-step answer:
Let there be two boys A and B.
As we know that there are ten boys.
Now as we know that according to the identity of permutation which states that if there is are r persons and total n identical items and all n items are distributed to r persons and let each of the \[{i^{th}}\] person is given \[{m_i}\] balls then the number of ways for this distribution is given as \[\dfrac{{n!}}{{\left( {{m_1}!} \right) \times \left( {{m_2}!} \right) \times ..... \times \left( {{m_r}!} \right)}}\]
And we know that if r is any positive integer than \[r! = r \times \left( {r - 1} \right) \times \left( {r - 2} \right) \times ..... \times 2 \times 1\]
Now as we know that two persons are given 2 and 8 balls.
So, there can be two possible cases.
Case 1:-
Person A is given 2 balls and person B is given 8 balls
So, number of ways for this distribution will be \[\dfrac{{10!}}{{\left( {2!} \right) \times \left( {8!} \right)}} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2 \times 1} \right) \times \left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)}} = 45\]
Case 2:-
Person A is given 8 balls and person B is given two balls.
So, number of ways for this distribution will be \[\dfrac{{10!}}{{\left( {8!} \right) \times \left( {2!} \right)}} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) \times \left( {2 \times 1} \right)}} = 45\]
So, the total number of ways to distribute 10 balls such that one person gets 2 and the other gets 8 balls will be equal to the sum of ways for both cases.
Hence, the total number of ways will be 45 + 45 = 90.
Note: Whenever we come up with this type of problem then if all objects are identical and it is not specified that which person is given particular amount of objects then there can be many cases like any of the person can get the specified amount of objects (like here A can be given 8 or 2 balls and B can also be given 8 or 2 balls). So, we have to divide the total ways into many cases and then find the number of possible ways for each case and after that find the sum of ways of all cases to get the total possible ways. This will be the efficient way to find the solution of the problem.
Complete step-by-step answer:
Let there be two boys A and B.
As we know that there are ten boys.
Now as we know that according to the identity of permutation which states that if there is are r persons and total n identical items and all n items are distributed to r persons and let each of the \[{i^{th}}\] person is given \[{m_i}\] balls then the number of ways for this distribution is given as \[\dfrac{{n!}}{{\left( {{m_1}!} \right) \times \left( {{m_2}!} \right) \times ..... \times \left( {{m_r}!} \right)}}\]
And we know that if r is any positive integer than \[r! = r \times \left( {r - 1} \right) \times \left( {r - 2} \right) \times ..... \times 2 \times 1\]
Now as we know that two persons are given 2 and 8 balls.
So, there can be two possible cases.
Case 1:-
Person A is given 2 balls and person B is given 8 balls
So, number of ways for this distribution will be \[\dfrac{{10!}}{{\left( {2!} \right) \times \left( {8!} \right)}} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2 \times 1} \right) \times \left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right)}} = 45\]
Case 2:-
Person A is given 8 balls and person B is given two balls.
So, number of ways for this distribution will be \[\dfrac{{10!}}{{\left( {8!} \right) \times \left( {2!} \right)}} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) \times \left( {2 \times 1} \right)}} = 45\]
So, the total number of ways to distribute 10 balls such that one person gets 2 and the other gets 8 balls will be equal to the sum of ways for both cases.
Hence, the total number of ways will be 45 + 45 = 90.
Note: Whenever we come up with this type of problem then if all objects are identical and it is not specified that which person is given particular amount of objects then there can be many cases like any of the person can get the specified amount of objects (like here A can be given 8 or 2 balls and B can also be given 8 or 2 balls). So, we have to divide the total ways into many cases and then find the number of possible ways for each case and after that find the sum of ways of all cases to get the total possible ways. This will be the efficient way to find the solution of the problem.
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