Question

In how many ways 13 non distinguishable books can be distributed among 7 persons so that every person gets at least one book and at least one person gets 4 books but not more.

Answer 1

In this particular type of question use the concept that if we have to choose r objects out of n objects then the number of ways of choosing is ${}^n{C_r}$ and construct the cases according to given information and the final answer is the sum of these cases so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given data:
There are 13 non distinguishable books.
Now we have to distribute these books to 7 people in such a way that every person gets at least one book and at least one person gets 4 books but not more.
So the possible number of cases which are arise are given below:
Case – 1: when we give 4 books to one person, 2 books to three people and one book each to the remaining persons.
So the number of ways to select one person out of 7 which we give 4 books = ${}^7{C_1}$.
Now the remaining persons are (7 - 1) = 6.
So the number of ways to select 3 persons out of 6 which we give 2 books each = ${}^6{C_3}$
Now the remaining persons = (6 - 3) = 3
So the number of ways to select 3 persons out of 3 which we give 1 book each = ${}^3{C_3}$
So the number of ways to give 4 books to one person, 2 books to three person and one books to remaining person is the multiplication of the above calculated values = ${}^7{C_1} \times {}^6{C_3} \times {}^3{C_3}$
Case – 2: when we give 4 books to one person, 3 books to one person, 2 books to one person and 1 book each to the remaining persons.
So the number of ways to select one person out of 7 which we give 4 books = ${}^7{C_1}$.
Now the remaining persons are (7 - 1) = 6.
So the number of ways to select 1 person out of 6 which we give 3 books = ${}^6{C_1}$
Now the remaining persons = (6 - 1) = 5
So the number of ways to select 1 person out of 5 which we give 2 books each = ${}^5{C_1}$
Now the remaining persons = (5 - 1) = 4
So the number of ways to select 4 persons out of 4 which we give 1 book each = ${}^4{C_4}$
So the number of ways to give 4 books to one person, 3 books to one person, 2 books to one person and 1 book each to remaining persons is the multiplication of the above calculated values = ${}^7{C_1} \times {}^6{C_1} \times {}^5{C_1} \times {}^4{C_4}$
Case – 3: when 4 books each to two persons and one book each to remaining persons
So the number of ways to select two persons out of 7 which we give 4 books = ${}^7{C_2}$.
Now the remaining persons are (7 - 2) = 5.
So the number of ways to select 5 persons out of 5 which we give 1 book each = ${}^5{C_5}$
So the number of ways to give 4 books each to two persons and one boo each to remaining persons is the multiplication of the above calculated values = ${}^7{C_2} \times {}^5{C_5}$
Now the total number of ways is the addition of the above cases.
So the total number of ways to distribute 13 non distinguishable books can be distributed among 7 persons so that every person get at least one book and at least one person gets 4 books but not more is,
= ${}^7{C_1} \times {}^6{C_3} \times {}^3{C_3}$ + ${}^7{C_1} \times {}^6{C_1} \times {}^5{C_1} \times {}^4{C_4}$ + ${}^7{C_2} \times {}^5{C_5}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so we have,
= $7 \times \dfrac{{6.5.4}}{{3.2.1}} \times 1 + 7 \times 6 \times 5 \times 1 + \dfrac{{7.6}}{{2.1}} \times 1$
$\Rightarrow 7 \times 20 \times 1 + 42 \times 5 + 21 \times 1$
$\Rightarrow 140 + 210 + 21 = 371$
So this is the required answer.

Note :Whenever we face such types of questions the key concept we have to remember is that always recall the formula of combination which is stated above, then evaluate all the cases as above evaluated then add all these cases as above and simplify using the combination property we will get the required answer.