Question

In how many ways can a man send three of his children to seve different colleges of a certain town?

please help me
solve that ​

Answer 1

Answer:

Let S1, S2, S3, S4, S5, and S6 are the six schools and A, B, and C are the three sons.

We do not know which two sons are to be in the same school. If we know that then the answer will be different. Anyway I will try solve both ways:

Two sons can be selected in 3C2 or 3 ways.

They can go to a particular school in 6 ways. That makes it 18 ways. The third son can go to his school in 5 ways. Altogether, we have 18*5 = 90 ways.

If we know the particular two sons, then these two can go to one particular school in 6 ways. The third son can select any of the five remaining schools. That makes it 30 ways.

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Answer 2

Given,

The number of children$\[ = 3\]$

The number of colleges$\[ = 7\]$

Solution,

Know that two sons are to be in the same school.

So, the number of ways

$\[\begin{array}{l}{ \Rightarrow ^3}{C_2}\\ \Rightarrow 3\end{array}\]$  

Understand that out of these two can go to one particular school in 6 ways. The third son can select any of the five remaining schools.

Therefore, the total number of ways is $\[30.\]$

Hence, the number of ways are $\[30.\]$