Combinations problem

It's like selecting three things out of 6 available things.

So, No. of ways = 6C3 = 6!/3!*3! = 20

Lets say flavour are A,B,C,D,E, AND F.

Start selecting three flavour ---

ABC
ABD
ABE
ABF....You will get a total of 20 flavour.

remeber whenever u r choosing from a bigger lot where the order doesnt matter, one uses combination...
=> nCr = n!/r!(n-r)!
so here u have a total of 6 flavours and u have to choose 3...
so 6C3= 20

radhika108 wrote:there are six flavors of ice cream: banana, chocolate, lemon, strawberry and vanilla and pistachio. You can have three scoops. How many variations will there be?. It has been worked out as(r+n-1)!/r!(n-1)! Can somebody explain the working for this. thanks in advance.

This is problem of selecting items from a set where the order does not matter and at the same time repetition is allowed.
n=6, r=3

Here we know that we have to select 3 flavors. But how many options do we have for selection? Is it 6? NO. B'cos repetition is allowed.

Imagine, you are at an ice-cream counter and you can take 3 scoops from 6 flavors available. Note that having selected a flavor you can still stand there and select it again b'cos repetition is allowed. You always move from left to right and once you say No (N) to one flavor, you move on to next flavor.

So, the counter from left to right is like (banana, chocolate, lemon, strawberry, vanilla, pistachio)

Let say one selects (banana, vanilla, pistachio): that is YNNNNYNY where Y corresponds to selecting a flavor and N corresponds to not selecting a flavor.
Lets say one selects (banana, lemon, vanilla): that is YNNYNNYN
Lets say one selects (banana, chocolate, lemon): that is YNYNYNNN
Let's say one selects (banana, banana, banana): that is YYYNNNNN

Note that once you are done with 3 scoops, you can't say NO to that same flavor anymore b'cos no one is asking you to take more You have to just move ahead saying NO politely to the remaining options

So, we have to say yes 3 times and No 5 times. That is we have 8 options. That is 6+3-1.

So, the answer is (r+n-1)C(r) = 56 ways.


Further: n=6,r=3
If order matters and repetition allowed
6*6*6= 6^3 = 216 ways
If order matters and repetition not allowed
6P3 = 6!/3! =120 ways
If order does not matter and repetition not allowed
6C3= 6!/(3!*3!) = 20 ways

HTH!!

thanks iamseerforthe answer. It has been explained as xxx and os or |and os or YN . This is a different way(technique) to think of the problem,which is not very intuitive to me. Is there a more direct way that has a logical flow, eg, for somebody who is not aware of this technique.

iamseer wrote:

radhika108 wrote:there are six flavors of ice cream: banana, chocolate, lemon, strawberry and vanilla and pistachio. You can have three scoops. How many variations will there be?. It has been worked out as(r+n-1)!/r!(n-1)! Can somebody explain the working for this. thanks in advance.

This is problem of selecting items from a set where the order does not matter and at the same time repetition is allowed.
n=6, r=3

Here we know that we have to select 3 flavors. But how many options do we have for selection? Is it 6? NO. B'cos repetition is allowed.

Imagine, you are at an ice-cream counter and you can take 3 scoops from 6 flavors available. Note that having selected a flavor you can still stand there and select it again b'cos repetition is allowed. You always move from left to right and once you say No (N) to one flavor, you move on to next flavor.

So, the counter from left to right is like (banana, chocolate, lemon, strawberry, vanilla, pistachio)

Let say one selects (banana, vanilla, pistachio): that is YNNNNYNY where Y corresponds to selecting a flavor and N corresponds to not selecting a flavor.
Lets say one selects (banana, lemon, vanilla): that is YNNYNNYN
Lets say one selects (banana, chocolate, lemon): that is YNYNYNNN
Let's say one selects (banana, banana, banana): that is YYYNNNNN

Note that once you are done with 3 scoops, you can't say NO to that same flavor anymore b'cos no one is asking you to take more You have to just move ahead saying NO politely to the remaining options

So, we have to say yes 3 times and No 5 times. That is we have 8 options. That is 6+3-1.

So, the answer is (r+n-1)C(r) = 56 ways.


Further: n=6,r=3
If order matters and repetition allowed
6*6*6= 6^3 = 216 ways
If order matters and repetition not allowed
6P3 = 6!/3! =120 ways
If order does not matter and repetition not allowed
6C3= 6!/(3!*3!) = 20 ways

HTH!!

thanks iamseerforthe answer. It has been explained as xxx and os or |and os or YN . This is a different way(technique) to think of the problem,which is not very intuitive to me. Is there a more direct way that has a logical flow, eg, for somebody who is not aware of this technique.

I know the method is seemingly counter-intuitive. It's like when repetition is allowed and arrangement order does not matter, we have to select r and skip (n-1)

Can't think of a better way to explain that to you at the moment. It is basically about counting methods. I'll think and let you know if I come up with a method that you would possibly find intuitive.

In the mean while, why don't you try and work out selecting 2 objects from 4 the crude way. That is by jotting down all possibilities:

Case 1: Order matters and repetition allowed (you should get 4*4=16 ways)
Case 2: Order matters and repetition not allowed (you should get 4!/2! = 12 ways)
Case 3: Order does not matter and repetition not allowed (you should get 4!/(2!*2!) = 6 ways)
Case 4: Order does not matter and repetition allowed (you should get 5!/(3!*2!)= 10 ways)

HTH!!
Thanks.