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In how many ways $12$ cars can be divided into $3$ groups such that all contains $4$ cars to each child.
Option: 1
$57700$

Option: 2
$57750$

Option: 3
$495$

Option: 4
$5775$

Answers (1)

best_answer

The given information is:

Number of cars $=2$

Number of groups $=3$

Number of cars to each groups $=4$

Pick first $4$ cars from $12$ cars,

Total no. of ways of choosing is:

$\frac{12 !}{(12-4) ! 4 !}=\frac{12 !}{8 ! 4 !}=495$

Pick next $4$ cars from $8$ cars,

Total no. of ways of choosing is:

$\frac{8 !}{(8-4) ! 4 !}=\frac{8 !}{4 ! 4 !}=70$

Pick last $4$ cars from $4$ cars and the number of ways this can be done is $1$ .

Since the three teams are similar, there is no differentiation between them. Therefore, we need to divide it with $3 !=6$ .

The total numbers of ways:

$\frac{495 \times 70}{6}=5775$

Posted by

avinash.dongre

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