MHB Find N(7,4)/N(7,5): Find the Answer Here

[evinda]

Hi! :D
In Lotto,there are the numbers $L=\{ 1,...,49 \}$.A Lotto ticket with $k$ options is a subset of $L$ with $k$ elements.$6$ numbers are selected in a draw.Given a draw, let $N(k,m)$ the number of the different tickets with $k$ options,that have $m$ successes $( 6 \leq k \leq 49$ and $ 0 \leq m \leq 6)$.Then, $ \frac{N(7,4)}{N(7,5)}$ is equal to :

• $\frac{205}{6}$
  
• $\frac{43}{2}$
  
• $\frac{215}{12}$

Could you help me to find which of the above answers is right?

 

[I like Serena]

Hi! :D
In Lotto,there are the numbers $L=\{ 1,...,49 \}$.A Lotto ticket with $k$ options is a subset of $L$ with $k$ elements.$6$ numbers are selected in a draw.Given a draw, let $N(k,m)$ the number of the different tickets with $k$ options,that have $m$ successes $( 6 \leq k \leq 49$ and $ 0 \leq m \leq 6)$.Then, $ \frac{N(7,4)}{N(7,5)}$ is equal to :

• $\frac{205}{6}$
  
• $\frac{43}{2}$
  
• $\frac{215}{12}$

Could you help me to find which of the above answers is right?

Hey! :)

Suppose the lotto numbers are always 1,2,3,4,5,6.
We can assume this without loss of generality.

Let's take a look at N(7,4).

Suppose the first 4 of our 7 numbers are in those 6 lucky numbers, and the remainder is not.
Then the number of successful combinations is:
$$6 \cdot 5 \cdot 4 \cdot 3 \cdot 43 \cdot 42 \cdot 41$$

But we can also have a different ordering.
There are $\binom 7 4$ ways to pick 4 out of 7 options.

So the total number is:
$$N(7,4) = \binom 7 4 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 43 \cdot 42 \cdot 41$$

 

[evinda]

Hey! :)

Suppose the lotto numbers are always 1,2,3,4,5,6.
We can assume this without loss of generality.

Let's take a look at N(7,4).

Suppose the first 4 of our 7 numbers are in those 6 lucky numbers, and the remainder is not.
Then the number of successful combinations is:
$$6 \cdot 5 \cdot 4 \cdot 3 \cdot 43 \cdot 42 \cdot 41$$

But we can also have a different ordering.
There are $\binom 7 4$ ways to pick 4 out of 7 options.

So the total number is:
$$N(7,4) = \binom 7 4 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 43 \cdot 42 \cdot 41$$

I haven't understood how you found $N(k,m)$.. (Worried) Could you explain it to me?? (Thinking)(Thinking)(Thinking)

 

[I like Serena]

I haven't understood how you found $N(k,m)$.. (Worried) Could you explain it to me?? (Thinking)(Thinking)(Thinking)

Can you explain what you do and do not understand?
What are your thoughts? (Wondering)(Thinking)

 

[evinda]

Can you explain what you do and do not understand?
What are your thoughts? (Wondering)(Thinking)

I have understood how we found $\binom 7 4$.It means that from $7$ options,the $4$ are successes..
But how did you find this:
$ 6 \cdot 5 \cdot 4 \cdot 3 \cdot 43 \cdot 42 \cdot 41$ ?

Why do we have to multiplicate $\binom 7 4$ with this product?? (Worried)(Thinking)

 

[I like Serena]

I have understood how we found $\binom 7 4$.It means that from $7$ options,the $4$ are successes..
But how did you find this:
$ 6 \cdot 5 \cdot 4 \cdot 3 \cdot 43 \cdot 42 \cdot 41$ ?

Why do we have to multiplicate $\binom 7 4$ with this product?? (Worried)(Thinking)

We're assuming the first 4 choices are successes.
Since there are 6 successful numbers, the first choice must be one of those 6.

Since the second choice is also a success, it must be one of the remaining 5 successful numbers (we've already used up 1 successful number).

In other words, there are $6 \cdot 5$ ways that the first choice is successful and that the second choice is also successful.

How would you continue?

 

[evinda]

We're assuming the first 4 choices are successes.
Since there are 6 successful numbers, the first choice must be one of those 6.

Since the second choice is also a success, it must be one of the remaining 5 successful numbers (we've already used up 1 successful number).

In other words, there are $6 \cdot 5$ ways that the first choice is successful and that the second choice is also successful.

How would you continue?

The third choice is also a success, it must be one of the remaining 4 successful numbers and the fourth is also a success, so it must be one of the remaining 3 successful numbers.
So,we have so far $6 \cdot 5 \cdot 4 \dot 3$ ways that the first 4 choises are successes.
Then,there are $43$ choices,that the next choice is not a success and so on..
Right? (Blush)

Couldn't we also do it like that?(Thinking)

$\binom{6}{m} \cdot \binom{49-6}{k-m}$