Using Small Examples to Solve Combinatorial Problems

[sahilmm15]

In the below picture I understood the problem and also understood the solution. But I am not clear on why did they choose this particular method. So my question is why did they divided 50 with powers of 2 and 3, and what relation does $50$ have with $50!$ ( I am clear on the solution part but I am clueless how would have the author thought of the problem.) Thanks!

 

[PeroK]

It's just counting factors isn't it? It's not clear how much you understand.

 

[sahilmm15]

It's just counting factors isn't it? It's not clear how much you understand.

I think I overcomplicated it. Can you explain me from scratch.

 

[PeroK]

I think I overcomplicated it. Can you explain me from scratch.

If you look at $10!$ say, then count the factors of $2$. We have:

2, 4, 6, 8 and 10 all have a factor of 2 (that's five factors of 2)

4 and 8 have a second factor of 2 (that's two more factors of 2)

8 has a third factor of 2 (that's another one)

Altogether, therefore, $10!$ has eight factors of 2. And doing this for the other primes we see that: $$10! = 2^8 \times 3^4 \times 5^2 \times 7$$

 

[sahilmm15]

If you look at $10!$ say, then count the factors of $2$. We have:

2, 4, 6, 8 and 10 all have a factor of 2 (that's five factors of 2)

4 and 8 have a second factor of 2 (that's two more factors of 2)

8 has a third factor of 2 (that's another one)

Altogether, therefore, $10!$ has eight factors of 2. And doing this for the other primes we see that: $$10! = 2^8 \times 3^4 \times 5^2 \times 7$$

This was so simple. Thanks!

 

[sahilmm15]

If you look at $10!$ say, then count the factors of $2$. We have:

2, 4, 6, 8 and 10 all have a factor of 2 (that's five factors of 2)

4 and 8 have a second factor of 2 (that's two more factors of 2)

8 has a third factor of 2 (that's another one)

Altogether, therefore, $10!$ has eight factors of 2. And doing this for the other primes we see that: $$10! = 2^8 \times 3^4 \times 5^2 \times 7$$

What I learned from this example is if you cannot comprehend a bigger problem take a smaller example and relate to the steps. I saw you doing this from past 2 answers and it worked beautifully.

 

[PeroK]

What I learned from this example is if you cannot comprehend a bigger problem take a smaller example and relate to the steps. I saw you doing this from past 2 answers and it worked beautifully.

Especially for these combinatorial problems. If in doubt, count for a low-value example. Often that's how you see what's going.

And, it gives you an answer to check against any general formula you come up with.