A difficult task

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

Joe Bartolozzi wrote:

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

4845 ways. Calculate C(20,4)

Claudio NVKP wrote:

5*2

Claudio NVKP wrote:

Joe Bartolozzi wrote:

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

4845 ways. Calculate C(20,4)

How so, common sense doesn't tell me that.

Say we have 'n' items and we have to select 'k' items from them, we use C(n,k)= n!/k!.(n-k)!

also, a sidenote n!= n*n-1*n-2*n-3.............*1

Joe Bartolozzi wrote:

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

4845 ways. Calculate C(20,4)

K.Rokossovski wrote:

Claudio NVKP wrote:

5*2

That is correct, though the actual answer should be: (5!/3!)/2!

K.Rokossovski wrote:

Joe Bartolozzi wrote:

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

4845 ways. Calculate C(20,4)

OK. Lets call the object A, B, C, D, and E. Now give me just six of those 4845 ways.

K.Rokossovski wrote:

Joe Bartolozzi wrote:

K.Rokossovski wrote:

No, there's only 5. There's only five different items that can be missing from the selected set.

4845 ways. Calculate C(20,4)

OK. Lets call the object A, B, C, D, and E. Now give me just six of those 4845 ways that's why i multiplied them. But I was blind for the otger question though.

I don't understand what you mean... but again, if you think there are more than 5, give me just 6 of them.

K.Rokossovski wrote:

Claudio NVKP wrote:

5*2

That is correct, though the actual answer should be: (5!/3!)/2!

Any 4th grade level math question would be about my theoretical capacity. Not above that though.

Claudio NVKP wrote:

K.Rokossovski wrote:

Claudio NVKP wrote:

5*2

That is correct, though the actual answer should be: (5!/3!)/2!

So 5! Is 5, 3! is 3, and 2! is 2. So it's 2(5/3)?

3!=1*2*3=6.

2! = 1*2=2.

So it is (120/6)/2 = 20/2 = 10

In normal language: you're only picking two objects; there's five ways to pick the first one, then 4 to pick the next. You're not picking the other three objects (which would be 5!), so you should divide by the ways you could have done those.

Then you have to eliminate duplicates, because the result currently has both AB and BA in it (depending on the pick order, but that is not relevant for the result set). How many ways are there for the order to pick two objects? well, 2!

So if you want to make it into a formula for different ways of picking x objects from a set of y items, the formula would be:

ways = (y!/(y-x)!)/x!

Well, you were supposed to pic 4 objects from 20

Lets say that the objects in the set of {A,B,C,.......,T} (20)

Fix object 'A,B,C' and make combinations around it, they are:

A,B,C,D

A.B,C,E

A,B,C,F

......

17 combinations made this way. Keep on doing this and obtain 4845 ways to do so.

17, but then it would get a bit lower and lower. Wouldn’t the next be like 16, 15, 14, and you add all of that? Wouldn’t come to 4000…

Carking the 6th wrote:

17, but then it would get a bit lower and lower. Wouldn’t the next be like 16, 15, 14, and you add all of that? Wouldn’t come to 4000…

Use C(20,4) for this.

To calculate the number of ways to select 4 items from a set of 20 distinct items, you can use the combination formula, which is denoted as "n choose k" and is calculated as:

Combination(n,k)=n!k!(n−k)!Combination(n,k)=k!(n−k)!n!

Where:

• n!n! denotes the factorial of nn, which is the product of all positive integers up to nn.
• k!k! denotes the factorial of kk.
• nn is the total number of items.
• kk is the number of items to be selected.

Combination(20,4)=20!4!(20−4)!Combination(20,4)=4!(20−4)!20!

Let's compute this:

Combination(20,4)=20×19×18×174×3×2×1Combination(20,4)=4×3×2×120×19×18×17Combination(20,4)=4845Combination(20,4)=4845

So, there are 4845 ways to select 4 items from a set of 20 distinct items.

Carking the 6th wrote:

To calculate the number of ways to select 4 items from a set of 20 distinct items, you can use the combination formula, which is denoted as "n choose k" and is calculated as:

Combination(n,k)=n!k!(n−k)!Combination(n,k)=k!(n−k)!n!

Where:

• n!n! denotes the factorial of nn, which is the product of all positive integers up to nn.
• k!k! denotes the factorial of kk.
• nn is the total number of items.
• kk is the number of items to be selected.

Plugging in the values, you get:

Combination(20,4)=20!4!(20−4)!Combination(20,4)=4!(20−4)!20!

Let's compute this:

Combination(20,4)=20×19×18×174×3×2×1Combination(20,4)=4×3×2×120×19×18×17Combination(20,4)=4845Combination(20,4)=4845

So, there are 4845 ways to select 4 items from a set of 20 distinct items.

Maths

answer this simple question:

how many numbers are there between 7775 and76542