Question

In how many ways can 5 chocolates be distributed among 3 people such that each person gets at least 1 chocolate?​

Answer 1

Given:

Number of people = 3

Number of chocolates= 5

TO Find:

In how many ways can 5 chocolates be distributed among 3 people such that each person gets at least 1 chocolate

Factorial notation :

The notation n! represents the product of first n natural numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as ‘n factorial’. 

Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !

For example,

1! = 1

2! = 1 x 2 = 2

3! = 1 x 2 x 3 = 6

4! = 1 x 2 x 3 x 4 = 24, which are the factors of the given number.

Formula Used:

$nC_{r}$ = $\frac{n!}{(n-r)r!}$

Distributing n things among r person so that any of them can get none, one , two or all = $^{(n+r-1)} C_{r-1}$

Solution:

According to the question, the number of people are 3

Each of them should get atleast 1 chocolate

The remaining chocolates are 5-3 = 2

By using the formula ,we get

Distributing 2 things among 3 person so that any of them can get none, one , two or all = $^{(2+3-1)} C_{3-1}$

                           = $^{4} C_{2}$

                           = $\frac{4!}{(4-2)2!}$

                           =$\frac{4!}{(2)2!}$

                           =$\frac{4*3*2*1}{(2)*2*1}$

                           = 6

Distributing 2 things among 3 person so that any of them can get none, one , two or all   =6

In 6  ways can 5 chocolates be distributed among 3 people such that each person gets at least 1 chocolate

Answer 2

Step 1: Given data

Number of chocolates$=5$

Number of people among which chocolates have to be distributed$=3$

Number of ways in which each person gets at least $1$ chocolate$=?$

Step 2: Using the formula

$_{}^{n}\textrm{C}_{r}=\frac{n!}{(n-r)!r!}$

To distribute $n$ things among $r$ people so that each of them gets zero, one, two or all of them is  given by,

$_{}^{n+r-1}\textrm{C}_{r-1}$

Step 3: Calculating the number of ways

If there are $3$ people, and each of them must have at least $1$ chocolate then,

Remaining chocolates are, $5-3=2$

We can distribute $2$ chocolates among $3$ people using,

$_{}^{n+r-1}\textrm{C}_{r-1}\\\\=_{}^{2+3-1}\textrm{C}_{3-1}\\\\ =_{}^{4}\textrm{C}_{2}\\\\ =_\frac{4!}{2!2!}= \frac{4\times 3\times 2\times 1}{2\times1 \times2 \times 1}\\\\=6$

Hence, the number of ways we can distribute $5$ chocolates to distribute amone $3$ people such that each of them gets at least $1$ chocolate is $6$ ways.

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