7 people are stopped randomly and asked about their birthdays

Question

7 persons are stopped on the road at random and asked about their birthdays.If the probability that 3 of them are born on Wednesday, 2 on Thursday and the remaining 2 on Sunday is $\frac {K}{7^6 }$ , then K is equal to

I tried solving using Bernoulli triads for each day and adding them together, but of no result.

Answer 1

This is like rolling a seven sided die $7$ times and getting $3$ twos, $2$ fives and $2$ sixes. There are $7^7$ possible outcomes and $\frac{7!}{2!2!3!}$ ways to get this particular outcome.

So $$P = \frac{\frac{7!}{2!2!3!}}{7^7}$$

But if $$P = \frac{K}{7^6} = \frac{\frac{7!}{2!2!3!}}{7^7}$$

Then $$K = \frac{\frac{7!}{2!2!3!}}{7^7}\cdot 7^6 = \frac{\frac{7!}{2!2!3!}}{7}$$ $$K = \frac{6!}{2!2!3!} = 30$$

Answer 2

Hint: In how many ways can you choose the $3$ people that are born on Wednesday? In how many ways can you choose the $2$ people born on Thursday? What can you finally say to deduce the answer?