Suppose that a large family has 14 children, including two sets of identical triplets, three sets of identical twins, and two individual children. How many ways are there to seat these children in a row of chairs if the identical triplets or twins cannot be distinguished from one another

Answer:

First, assume that we have 14 seats in a row.

Let's count the total number of combinations assuming that the 14 kids are differentiable.

In the first seat, we can sit only one child, the number of options is (14)

In the second seat, we can sit another child, in this case, the number of options is 13, because one child is already seated.

For the third seat, we will have 12 options.

for the fourth seat we will have 11 options, and so on.

The total number of permutations is equal to the product between the numbers of options for each case.

Then the total number of permutations is:

P = 14*13*12*11*...*3*2*1 = 14!

Now, we need to discount the permutations of the two sets of identical triplets:

a triplet has 3 elements, then the permutations of each set of triplets is:

p = 3*2*1 = 3!

Then if we discount two sets of triplets, we need to divide by:

d = 2*(3!)

We also have 3 sets of twins, the permutations for each set of twins is:

p = 2*1 = 2!

Then if we discount the 3 sets of twins, we need to by:

q = 3*(2!)

Then the total number of different combinations will be:

[tex]P = \frac{14!}{2*(3!)*3*(2!)} = 1,210,809,600[/tex]

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