Probability of Love...

Firstly, beautiful girls in engineering college?? You do know that it is not possible?

I'm kidding. Stereotype be still rampant among all my peers. Let's get to the question.

The total number of ways that the $5$ friends can fall in love is obviously $10^5$ as each person has $10$ choices.

Now, we need the number of cases where all of them fall for different persons. The way to do this is to choose five girls from the ten and pair them with each guy keeping in mind the order. This means we have $10C_5$ ways of choosing and $5!$ ways of pairing the chosen persons. Hence, the total number of favorable cases is $10C_5*5!$ . You can simply think of this as permutation of $10$ things taken $5$ at a time, i.e. $10P_5$ .

Therefore, the required probability is: $(10P_5)/10^5=189/625$

$189+625=\boxed{814}$

Total ways in which these $5$ boys can fall in love $= 10 \times 10 \times 10 \times 10 \times 10$

Total ways in which there $5$ boys fall in love such that their dream girls are different $= 10 \times 9 \times 8 \times 7 \times 6$

Probability that they don't clash over their dream girls $= \dfrac{10 \times 9 \times 8 \times 7 \times 6}{10 \times 10 \times 10 \times 10 \times 10} = \dfrac{9 \times 7 \times 3}{25 \times 25} = \dfrac{189}{625}$

From a probability perspective:

The first boy has probability $1$ of choosing an unchosen girl; the next, a $\tfrac{9}{10}$ probability; then $\tfrac{8}{10}$ , and so on. Each choice is independent, so:

$P = 1\cdot\tfrac{9}{10}\cdot\tfrac{8}{10}\cdot\tfrac{7}{10}\cdot\tfrac{6}{10} = \tfrac{189}{625}$ .

$189 + 625 = 814.$