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# Number of ways in which 9 different toys can be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is

Last updated date: 13th Jun 2024
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Hint:The distribution of toys are given in such a way that the toys of number 9 should be divided into the 4 children with three having even numbers of toys and one having three, meaning the number of toys given are in order of 2, 2, 2, 3. Hence, the combination should be done according to the toys divided into the three children in a formula of:
${}^{\text{Total number of toys}}{{C}_{\text{number of toys received by youngest}}}\dfrac{\left[ \text{Toys amongst even children} \right]!}{{{\left[ \text{Number of toys amongst even numbers!} \right]}^{\text{number of eldest}}}}$

Complete step by step solution:
According to the question given, the toys are divided amongst the eldest in the same number but one extra is given to the youngest, now the total number of toys are given as $9$.
So dividing the number of toys in terms of the number as given in the question as $2,2,2,3$.

The total number of toys amongst the eldest are divided is $2+2+2=6$.
Keeping the six places of the toys for the eldest separate, the youngest will get the total number of toys in a combination as:
${}^{\text{Total number of toys}}{{C}_{\text{number of toys received by youngest}}}$
Placing the values in the above formula we get the value of the combination as:
$\Rightarrow {}^{9}{{C}_{\text{3}}}$
Now the toys that were separated can be distributed in combinations of ${}^{\text{Total number of toys}}{{C}_{\text{number of toys received by youngest}}}\dfrac{\left[ \text{Toys amongst even children} \right]!}{{{\left[ \text{Number of toys amongst even numbers!} \right]}^{\text{number of eldest}}}}$
In which placing the values in the above formula, we get the combination as:
$\Rightarrow {}^{9}{{C}_{\text{3}}}\dfrac{6!}{{{\left( 2! \right)}^{3}}}$
This can be written as in another form which is:
$\Rightarrow \dfrac{9!}{3!{{\left( 2! \right)}^{3}}}$
Therefore, the total number of ways in which the toys can be distributed is given as $\dfrac{9!}{3!{{\left( 2! \right)}^{3}}}$.

Note:
We use the method of combination here because in combination, the method is used to select elements from a collection of various other elements whereas in permutation, the method is used to arrange the elements in total.