Problem of permutation and combination

Liuz · 2025-10-02 15:35:35

The school has newly purchased five different sets of experimental equipment,which are expected to be allocated to the laboratories of the first,second and third grades of high school. It is required that each grade be allocated at least one set of experimental equipment. How many ways can we allocate the equipments?

Bob · 2025-10-02 21:25:20

hi Luiz,

I'll have a go.

Firstly I'll ignore what kind of equipment is allocated and just work out the ways of distributing identical equipment

Every grade must have one, so I just need to consider what to do with the remaining two.

I think there are 6 options:

Ist grades get both; 2nd grades get both; 3rd grades get both;
Ist and 2nd get one each; 1st and 3rd get one each; 2nd and 3rd get one each.

Now to consider the different types of equipment. I can shuffle what is allocated in 5! ways.

So I think the answer to your question is 6 times 5! = 720 ways.

Bob

Liuz · 2025-10-02 22:18:45

Hi Bob,
Thanks for your help.
The 6 options inspired me.
I divided them into two groups:
One grade gets 3. For the other grades,each grade gets 1; one grade gets 1. For the other grades,each gets 2.
There are [(5!/2!)/3!]*2!*3=60 ways in the first group and [(5!/3!)/2!]*(3!/2!)*3=90 ways in the second group. So there are 150 ways in total.

Bob · 2025-10-03 01:41:15

I only get 30 for that first group

Calling the equipment types V W X Y Z and giving three to 1st grade we have

VWX Y Z
VWX Z Y
____________________
VWY X Z
VWY Z X
____________________
VWZ X Y
VWZ Y X
____________________
____________________
VXY W Z
VXY Z W
____________________
VXZ W Y
VXZ Y W
____________________
____________________
VYZ W X
VYZ X W
____________________
____________________
WXY V Z
WXY Z V
____________________
WXZ Z Y
WXZ Y Z

WYZ V X
WYZ X V
____________________
XYZ V W
XYZ W V
____________________

That makes 10 ways. Then multiply by 3 for 2nd and 3rd grades getting the three. 30 + 90 = 120

Bob

Liuz · 2025-10-03 11:43:54

Why not multiply 3!=6？

phrontister · 2025-10-03 16:05:20

Hi Bob;

Bob wrote:

That makes 10 ways.

Isn't each of your 20 lines a distinct permutation, making 20 ways?

If so, that yields 20 * 3 = 60...and 60 + 90 = 150.

Bob · 2025-10-03 20:10:50

Oh whoops!

I thought my answer was correct so I was looking for a way to show this. I started with the second part of Luiz's anaswer and showed 90 was correct. So then I tackled the first part and made that careless error. Very sorry.

The good news is I seem to have shown by 'exhaustion' that 150 is correct.

Bad news. I cannot see what is wrong with my analysis either. So I'm back to the 'drawing board' to find why.

Apologises for my excessive use of English metaphors.

Bob

Bob · 2025-10-03 20:30:55

OK I've got it.

If one grade gets 3, then there are 3 ways to achieve this so 3 x 5!. But some possibilities are repeated. I am counting vwx, vxw, wvx, wxv, xvw, xwv so I need to divide by 6.

If 2 grades get 2 each, again 3 ways, so 3 x 5!, but this time I'm counting vy and yv in one grade and wz and zw in a second grade so I need to divide by 2! and 2! again.

So I should have calculated thus:

Bob

Liuz · 2025-10-03 21:32:02

I’m truly grateful for your assistance. It turns out to be all efforts you made will pay back.

Math Is Fun Forum

#1 2025-10-02 15:35:35

Problem of permutation and combination

#2 2025-10-02 21:25:20

Re: Problem of permutation and combination

#3 2025-10-02 22:18:45

Re: Problem of permutation and combination

#4 2025-10-03 01:41:15

Re: Problem of permutation and combination

#5 2025-10-03 11:43:54

Re: Problem of permutation and combination

#6 2025-10-03 16:05:20

Re: Problem of permutation and combination

#7 2025-10-03 20:10:50

Re: Problem of permutation and combination

#8 2025-10-03 20:30:55

Re: Problem of permutation and combination

#9 2025-10-03 21:32:02

Re: Problem of permutation and combination

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