There are four identical copies each of five different magazines. In

Question

There are four identical copies each of five different magazines. In how many ways can the magazines be arranged in a row on a shelf, if nothing else is arranged with them?

(A) 5!/4!
(B) 5!/(4!)^5
(C) 20!/(5!)^4
(D) 20!/(4!)^5
(E) 20!/5(4!)

Vinit800HBS · Accepted Answer

Basically, the question says if we have 5 different magazines, say A B C D E, and each variety has 4 similar copies i.e. 
AAAA 
BBBB 
CCCC 
DDDD 
EEEE

then we can arrange them in

20! / (4! 4! 4! 4! 4!) i.e. 20!/(4!)^5

Option D

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Shrey9 · Answer

How do you get 20!, As the numerator..? I understand the denominator part well.

ParthSanghavi · Answer

Since you are now arranging the different types of magazines (A,B,C,D,E) which can be arranged in 5! ways. Hence shouldn't it be 5! * (4!)^5 ways??

Vinit800HBS · Answer

There are a total of 20 magazines. They all need to be arranged. Arranging 20 objects can be done in 20! Ways

Hope you understood.

autassumenda · Answer

Hi why isn't it 20! / (4!)^5 * 16! ?
I am confused is it because it falls under permutation?

Bunuel · Answer

So what’s the logic of multiplying by 16!? We have 20 magazines in total, 5 kinds, each repeated 4 times. Since all 20 are being arranged in a row, the total permutations are 20!. But because within each group of 4 identical magazines (like AAAA), those copies are indistinguishable, we divide by 4! for each of the 5 kinds. So the answer is 20!/(4!)^5. Multiplying by 16! makes no logical sense, nothing is being left unarranged or chosen.

autassumenda · Answer

Hi I was trying to follow the formula:

(n!)/((n-k)! * repetition!)

which in this case would lead to n-k = 16 
but I think I was on the completely wrong path, thanks

Bhavek · Answer

can you explain why option c is wrong

Bunuel · Answer

Option C is wrong because it uses the wrong numbers in the denominator.

There are 20 magazines in total (4 copies each of 5 types). For identical items, you divide by the factorial of the count of identical copies for  each  type. So you must divide by (4!) for each of the 5 kinds, giving (4!)^5.

Option C divides by (5!)^4, which assumes there are 4 types with 5 identical copies each, the situation is reversed. The correct setup is 20!/(4!)^5, not 20!/(5!)^4.

luisdicampo · Answer

Deconstructing the Question  
There are 5 different magazine types, with 4 identical copies of each.
Total magazines:
 5 \cdot 4 = 20

We must arrange them in a row, but copies of the same type are indistinguishable.

Step-by-step  
If all 20 magazines were distinct:
 20!

But for each type, the 4 identical copies can be rearranged internally in:
 4! 
ways without creating a new arrangement.

Since there are 5 types, divide by:
 (4!)^5

Total arrangements:
 \frac{20!}{(4!)^5}

Answer:     (D) 20!/(4!)^5

tryorcry · Answer

just for my understanding , 
4!x4!x4!x4!x4! = 4!^5 and doesnot equalsto 4!x5

There are four identical copies each of five different magazines. In

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