Question

In a preschool class of n, exactly n1 children are needed for activity 1, n2 for activity 2, and n3 for activity 3. Luckily n = n1 + n2 + n3. The teachers want to know in how many distinct ways the children can be assigned into these activities. (Two assignments are distinct if at least one student is in a different activity in each.) (a) They figure they could start by lining up the children arbitrarily. How many different line-ups are possible? (b) They then take the first n1 for activity 1, the next n2 for activity 2, and the rest for activity 3. How many different line-ups will create the exact same assignment of children to activities? (c) Use your answers from (a) and (b) to deduce the total number of distinct assignments.

Answer 1

Solution

Back-up Theory

Number of ways of arranging n distinct things among themselves (i.e., permutations)

= n!

= n(n - 1)(n - 2) …… 3.2.1……………………………………………………………….........................................................….….(1)

Complement of at least one is NONE and hence

Complement of at least one is different is ‘NONE is different’ or all are the same. ............................................................. (2)

Now, to work out the solution,

Part (a)

Every permutation of n children would yield a line up. So, vide (1),

Number of different line-ups possible is n! Answer 1

Part (b)

In the above count, every permutation of n₁ children in Activity would yield a line-up, but the children in the line-ups would be the same. Thus, {(n₁!) x (n₂!) x (n₃!)} line-ups would have the same children in each Activity. Hence,

Number of different line-ups creating the exact same assignment of children to activities is:

{(n₁!) x (n₂!) x (n₃!)} Answer 2

Part (c)

Vide (2), by complementary property,

the total number of distinct assignments = n1 - {(n₁!) x (n₂!) x (n₃!)} Answer 3

DONE