In how many of ways can 5 balls be placed in 4 tins if any

Question

In how many of ways can 5 balls be placed in 4 tins if any number of balls can be placed in any tin?

(A) 5C4
(B) 5P4
(C) 54
(D) 4^5
(E) 5^5

In this tricky problem why the answer D says 4^5 ??

we have 4 slots and any ball can be placed in any tin. hence, 4^4 at most.

Someone can axplain the reason behind the solution ??' I'm a little bit confused

thanks

Bunuel · Answer

Not a good question, we should know whether the balls are identical or not. From OA it seems that the question means that they are distinct. Now, each of the 5 different balls has 4 choices (4 tins to go to), so total # of distribution is 4*4*4*4*4=4^5. Answer: D. Check this for more: in-how-many-ways-can-5-different-rings-be-worn-in-four-126991.html Hope it helps.

Carcass · Answer

Sorry Bunuel I saw your links but at this point I need an explanation.

So, from the problem above (that I understood in somehow) I see that is the case of distribution. Infact we are talking of : DISTRIBUTION OF THINGS INTO GROUPS.

I see those concepts for the first time: i didn't see it neither in gmat math book nor in the four MGMAT guide (maybe I'm wrong).

I come to my question: are really important on the test ??' and we are talking of combination / permutation under different condition or a different concept of the fundamental principle of counting ??

Thanks.

himanshuhpr · Answer

@  
I am not getting the answer for this question
In this question if we can also consider following cases :

(1 CASE) Tin 1 (ALL 5 BALLS) + TIN 2 ( O BALLS)+ TIN 3 ( O BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/3! .

(2 CASE) Tin 1 ( 4 BALLS) + TIN 2 ( 1 BALLS)+ TIN 3 ( O BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/2! .

(3 CASE) Tin 1 ( 3 BALLS) + TIN 2 ( 2 BALLS)+ TIN 3 ( O BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/2! .

(4 CASE) Tin 1 ( 3 BALLS) + TIN 2 ( 1 BALLS)+ TIN 3 ( 1 BALLS) + TIN 4 ( O BALLS) --- TOTAL PERM-S.: 4!/2!

(5 CASE) Tin 1 ( 2 BALLS) + TIN 2 ( 1 BALLS)+ TIN 3 ( 1 BALLS) + TIN 4 ( 1 BALLS) --- TOTAL PERM-S.: 4!/3!

ADDING 1+2+3+4+5 = 4+12+12+12+4 =44 <<< WHATS WRONG IN THIS APPROACH ??

Strategist · Answer

From what I understand of a previous question, the problem with this approach is that it treats all the balls as equal. If all the balls are equivalent then
your approach is correct (I think) but imagine every ball was a different colour, there are suddenly a lot more combinations than 44.

There is another similar question here about putting letters in a box, if they are different letters then your approach doesn't work, but if they are equivalent (like junk mail) then your approach does work.

Maybe look that up.

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In how many of ways can 5 balls be placed in 4 tins if any

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