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Mathematics 44 Online
OpenStudy (anonymous):

Calculate the number of ways that 3 or 4 prizes can be awarded to a group of 5 people if: a) a member of the group is allowed to receive more than one prize b) a member of the group cannot receive more than one prize.

OpenStudy (anonymous):

May be begin with this now??

OpenStudy (anonymous):

*we..

OpenStudy (anonymous):

yes, sure.

OpenStudy (anonymous):

Firstly for A part..

OpenStudy (anonymous):

There are two conditions which says prizes can be 3 OR 4.. Right???

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

First let we have 3 prizes and 5 people... See, our very first a) part says that any member can be given any number of prizes.. Right???

OpenStudy (anonymous):

yep.

OpenStudy (anonymous):

So, suppose you are distributing 3 prizes.. There are 5 people.. you will give first prize in how many ways?? You can give it to first person or second or third like that.. First prize you can give in how many ways??

OpenStudy (anonymous):

5?

OpenStudy (anonymous):

How can JayDS be wrong?? Yes you are right.. Now note that carefully: any people can have any number of prizes.. So, suppose you at first gave me the the first prize.. Now second prize you can give in how many ways?? NOTE: You can still given me a Prize.. Getting?? Tell me the number of ways for second prize?

OpenStudy (anonymous):

lol, ur funny. uh so lets see, if I gave u the first prize then for the second prize there should be 5 ways.

OpenStudy (anonymous):

since I can still give u the second prize.

OpenStudy (anonymous):

Yes you are right.. Like wise third prize can be given in how many ways and I can still receive a prize..

OpenStudy (anonymous):

same, 5 ways.

OpenStudy (anonymous):

Yes.. So you can distribute 3 prizes to 5 people in \(5 \times 5 \times 5\) ways.. Right??

OpenStudy (anonymous):

yep.

OpenStudy (anonymous):

So, now if I assume there are 4 prizes instead of 3 then 4 prizes you can give in total how many ways?? Same procedure..

OpenStudy (anonymous):

yep, so 4x4x4 ways?

OpenStudy (anonymous):

oh wait.

OpenStudy (anonymous):

No...

OpenStudy (anonymous):

exactly the same?

OpenStudy (anonymous):

5x5x5 ways?

OpenStudy (anonymous):

There are 4 prizes and 5 people.. just close..

OpenStudy (anonymous):

See the ways you said are for 3 prizes where is the fourth prize gone??

OpenStudy (anonymous):

yeh, I don't know lol.

OpenStudy (anonymous):

I get confused easily, sorry.

OpenStudy (anonymous):

First prize you can give in 5 ways, Second prize you can give in 5 ways, Third prize you can give in 5 ways, Fourth prize you can give in 5 ways.. Now tell me what are the total number of ways??

OpenStudy (anonymous):

20 ways

OpenStudy (anonymous):

You have to multiply here.. I have multiplied it in case of 3 prizes 5*5*5... So now tell me??

OpenStudy (anonymous):

I think you are not getting what I said.

OpenStudy (anonymous):

in the case of 4 prizes it will still be 5*5*5=125 ways right? so the total would be 125+125 = 250 ways?

OpenStudy (anonymous):

No.. See, We have separately calculated for 3 prizes that is 5*5*5 Now we are calculating it for 4 prizes.. You have to just multiply them.. Aren't the number of ways to give 4 prizes 5*5*5*5 ??

OpenStudy (anonymous):

ok so u just times 5^4 but why do we do this?

OpenStudy (anonymous):

What?? Why we do multiplication instead of adding??

OpenStudy (anonymous):

um, why do we multiply 5x5x5x5? I know it's to find the number of ways to give 4 prizes but I don't quite understand it.

OpenStudy (anonymous):

See, we find it separately.. First prize can be given in 5 ways.. Second in 5 third in 5 and finally fourth in 5.. Right??

OpenStudy (anonymous):

yep.

OpenStudy (anonymous):

Now do you know what Counting Principle says??

OpenStudy (anonymous):

nope.

OpenStudy (anonymous):

Counting Principle says that: If there are two jobs such that one of them can be done in \(m\) ways and other can be done in \(n\) ways then, two jobs together can be completed in \(m \times n \) ways..

OpenStudy (anonymous):

ok.

OpenStudy (anonymous):

Similarly, here: If first prize can be distributed in 5 ways, Second prize can be in 5 ways, Third can be in 5 ways and Fourth can be in 5 ways, Then the four prizes in succession can be given in \(5 \times 5 \times 5\times5\) ways.. Not is it clear to you or not?

OpenStudy (anonymous):

*Now..

OpenStudy (anonymous):

ok, I believe I understand it now.

OpenStudy (anonymous):

So tell me after multiplying what you get in both the cases?? When you distribute 3 prizes how many ways you get?? When you gave 4 then how many ways you get..??

OpenStudy (anonymous):

5x5x5=125 5x5x5x5=625 125+625=750 ways

OpenStudy (anonymous):

Yes now you are right..

OpenStudy (anonymous):

If you are clear with this can we proceed for B?? Now we have to do the same but little bit different..

OpenStudy (anonymous):

yep, sure.

OpenStudy (anonymous):

I believe we will have to use factorial here in part b.

OpenStudy (anonymous):

B> part says that one cannot have prize more than one.. Right??

OpenStudy (anonymous):

yep.

OpenStudy (anonymous):

No need to use factorial, Simple multiplication will serve the purpose..

OpenStudy (anonymous):

kk.

OpenStudy (anonymous):

First is the case with having 3 prizes.. Out of these 3, first prize you can give in how many ways??

OpenStudy (anonymous):

three

OpenStudy (anonymous):

No dear.. There are 5 people in total.. you can give it to first, second, third fourth or fifth.. So, how many ways??

OpenStudy (anonymous):

oh yeh ==, 5.

OpenStudy (anonymous):

Suppose you gave me the first prize.. You cannot give me the second prize now.. So second prize can be given in how many ways??

OpenStudy (anonymous):

4.

OpenStudy (anonymous):

Yes, now the last prize you can given in how many ways??

OpenStudy (anonymous):

3

OpenStudy (anonymous):

Yes good.. Now apply the counting principle here.. Total how many ways you get?

OpenStudy (anonymous):

5x4x3=60 ways.

OpenStudy (anonymous):

and so for the 4 prizes it will be 5x4x3x2=120 then?

OpenStudy (anonymous):

You have really saved me.. Thanks.. Yes you are absolutely correct.. \(\large \color{green}\checkmark\)

OpenStudy (anonymous):

So total how many ways for second (B) part??

OpenStudy (anonymous):

180

OpenStudy (anonymous):

Any doubt, anywhere??

OpenStudy (anonymous):

lol, I'm really sorry! I am in fact a slow learner.

OpenStudy (anonymous):

If you are a slow leaner then it very good.. Always in a race of turtle and rabbit, turtle wins..

OpenStudy (anonymous):

yep, my only doubt is that I do understand it now, but I might forget how to do it properly in the test or other similar questions haha.

OpenStudy (anonymous):

and thanks for the positive comment, may I ask what is your name by the way? I would like to address you properly.

OpenStudy (anonymous):

I say honestly I am not good at Permutations And Combinations that is why I am saying that you should first listen to the lecture of your teacher.. He can explain you far better than me..

OpenStudy (anonymous):

You can call me Water.. My real name is Gurpreet...

OpenStudy (anonymous):

Yahoo is calling me I have to go now to help him.. Take care JayDS..

OpenStudy (anonymous):

yep, I understand. and I am actually still in high school so I just have classes not lectures yet and I find that you explain quite well considering it's text that I am reading and not like me listening in class.

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