There are five colleges in a city. In how many ways can a man send 3 of his sons to a college, if no two of his sons are to read in the same college ?

Answer

To solve the problem of how many ways a man can send 3 of his sons to 5 different colleges, ensuring that no two sons attend the same college, we can follow these steps:

### Step 1: Choose 3 Colleges from 5
First, we need to select 3 colleges out of the 5 available colleges. The number of ways to choose 3 colleges from 5 can be calculated using the combination formula:

\[ \text{Number of ways to choose 3 colleges} = \binom{5}{3} \]

Calculating this gives:

\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 2: Assign Sons to Selected Colleges
After selecting 3 colleges, we need to assign the 3 sons to these colleges. Since no two sons can attend the same college, we can assign them in different ways. The number of ways to assign 3 sons to 3 colleges is given by the number of permutations of 3 items, which is:

\[ \text{Number of ways to assign 3 sons} = 3! = 6 \]

### Step 3: Calculate Total Arrangements
Finally, to find the total number of ways to send 3 sons to 5 colleges, we multiply the number of ways to choose the colleges by the number of ways to assign the sons:

\[ \text{Total ways} = \binom{5}{3} \times 3! = 10 \times 6 = 60 \]

Thus, the total number of ways the man can send 3 of his sons to a college, ensuring that no two sons are in the same college, is **60**.

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