Question

There are three separate bundles of reading material comprising 4 comics, 2 novels and 3 magazines. They are placed together to form one pile.

a) In how many ways can this be done if there are no restrictions on where individual items are to be placed?

b) Determine the number of permutations if the order of the comic books in each bundle does not change.

Answer 1

as for a) it is just a simple factorial. if you combine them then there would be 4 + 2 + 3 = 10 total thingies.

the number of ways you can rearrange this is therefore 10!

as for b) it says permutation so you know you're going to use \\[nPr = \frac{n!}{r!}\]
it says permutation of comic books. there are 4 comics and there are 10 total thingies. there fore the permutation will be \[10P4 = \frac{10!}{4!}\]
got it?

Answer 2

@lgbasallote Be careful..

Answer 3

lol

Answer 4

b is wrong.

Answer 5

Seriously...

Answer 6

uhmm ookps yeah

Answer 7

got the formula wrong

Answer 8

\[nPr = \frac{n!}{(n-r)!}\]

Answer 9

Yes.......

Answer 10

sorry for the misinfo heh

Answer 11

\[10P4 = \frac{10!}{(10-4)!}\]

Answer 12

nice eye @waterineyes

Answer 13

Ha ha ha..

Answer 14

do you get that @JayDS ?

Answer 15

yep but the answer for b in my textbook is 1728 for some reason?

Answer 16

Do you have answers with you??

Answer 17

yep.

Answer 18

but sometimes the answers are wrong in the textbook, I'm not too sure.

Answer 19

hmm

Answer 20

don't worry, I'll ask my teacher later sometime, but thanks for the help.

Answer 21

But I think your textbook is right..

Because when we solved for wrong answers they we always say book can be wrong..

Ha ha ha..

Answer 22

so what's right @waterineyes eh?

Answer 23

lol, i'm not sure...

Answer 24

I am thinking lgba about the question..

Answer 25

What is the answer for a part @JayDS

Answer 26

10!=3628800

Answer 27

i doubt i did anything wrong :/

Answer 28

sorry got to eat dinner, afk.

Answer 29

Right.........

Answer 30

Are you there @JayDS ???

Answer 31

yep.

Answer 32

I am in doubt but I can show you how your book got 1728 as the answer for b part..

Answer 33

okok.

Answer 34

See firstly tell me :

Suppose, you have 3 things so In my how many you can arrange them ???

Answer 35

3!

Answer 36

*my will not come there..

Answer 37

Yes:

So you can arrange them in 6 ways..

Right??

Answer 38

yes

Answer 39

So write this 6 in your copy separately...

So out of these three things:

First is 2 novels

Second is 4 comics

and the third thing is 3 Magazines..

Right??

Answer 40

yep.

Answer 41

So now tell me in how many ways you can arrange 2 novels..??


In how many ways you will arrange 4 comics.??

In how many ways you will arrange 3 magazines??

Tell me separately the answers for all?

Answer 42

novels=2!= 2ways
comics=4!=24 ways
magazines=3!=6 ways
total of 32 ways.

Answer 43

I said you to use Counting Principle..

They will get multiply and not add..

Answer 44

2! x 4! x 3! =288 ways

Answer 45

Yes now remember our first answer..???

Answer 46

ya.

Answer 47

That you have written separately what was that??

Answer 48

wait igba's answer or...?

Answer 49

My answer..

Answer 50

I told you to separately write in your copy...

Answer 51

6

Answer 52

Yes..
Now multiply that with 288 ways and check what you got..

Answer 53

1728

Answer 54

Is this the answer you textbook says??

Answer 55

yes.

Answer 56

um, can u help me with some other questions?

Answer 57

May be..
I told you I am not good at this...

Can you ask your teacher something that I will say..

Answer 58

yep, no problem.

Answer 59

Question says that order of 4 comics does not change so there is no need to arrange 4 comics:

So.

6*6*2*4 = 288

So I think answer should be this...

So I am in doubt..

But according to your book's answer we have done it correctly..

Answer 60

ok.

Answer 61

Or most probably the answer will be:

\[= \frac{1728}{24} = 72\]

I said I am in doubt so be sure to get it confirmed from your teacher..

Answer 62

yep.