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Hi, suppose we want to arrange the letters "B,O,O,K", with replacement allowed (I.e. BBBB, OOOO etc..)

One method is as follows: $$\frac{4^4}{2!}$$  (Four slots, each slot can be any of the four letters so 4*4*4*4 , and then divide by 2! to remove the repitition of O's) giving 128 as an answer.

Another method: $$(3C1)^4=81$$  (Well, we have the distinct letters "B,O,K" , suppose we choose one letter for the first, there are 3C1 ways to do that, then same goes for the second etc... so 3C1*3C1*3C1*3C1 )

The second method gives the wrong answer...

Question: Why?

Apr 17, 2022
edited by Guest  Apr 17, 2022

#1
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Actually, the logic of the second one is the correct method..

Consider the simpler case: "AAB"

"How many 4 letter words can be formed using these three letters"

Well:

By listing:
AAAA
AAAB
AABB
ABBB
BBBB
BBBA
BBAA
BAAA
AABA
ABAA
ABAB
ABBA
BABA
BAAB
BABB
BBAB

Which yields 16..

The first method will give: $$\frac{3*3*3*3}{2!}$$ = 40.5 which is NOT an integer; hence this method must be wrong!

But the second method gives: $$2C1*2C1*2C1*2C1=2^4=16$$ which is the correct answer.....

It seems that the general technique is given for example "AAABBBCCCDDDDDEE" letters and we want to form "x-letter" word, then we make our set as: (A,B,C,D,E) and then we proceed by choosing 5*5*5*... (x times).

Please if anyone can help whether the above logic is correct or more original question.. or is there a better explanation?

Apr 17, 2022
#2
+1

B O O K ==4 letters.

Your first method is wrong NOT the second! Take the two "OO" as ONE letter and you have:

3^4 ==81 permutations. The 2nd method is correct.

So, the 81 permutations look like this:

{{B, B, B, B}, {B, B, B, K}, {B, B, B, O}, {B, B, K, B}, {B, B, K, K}, {B, B, K, O}, {B, B, O, B}, {B, B, O, K}, {B, B, O, O}, {B, K, B, B}, {B, K, B, K}, {B, K, B, O}, {B, K, K, B}, {B, K, K, K}, {B, K, K, O}, {B, K, O, B}, {B, K, O, K}, {B, K, O, O}, {B, O, B, B}, {B, O, B, K}, {B, O, B, O}, {B, O, K, B}, {B, O, K, K}, {B, O, K, O}, {B, O, O, B}, {B, O, O, K}, {B, O, O, O}, {K, B, B, B}, {K, B, B, K}, {K, B, B, O}, {K, B, K, B}, {K, B, K, K}, {K, B, K, O}, {K, B, O, B}, {K, B, O, K}, {K, B, O, O}, {K, K, B, B}, {K, K, B, K}, {K, K, B, O}, {K, K, K, B}, {K, K, K, K}, {K, K, K, O}, {K, K, O, B}, {K, K, O, K}, {K, K, O, O}, {K, O, B, B}, {K, O, B, K}, {K, O, B, O}, {K, O, K, B}, {K, O, K, K}, {K, O, K, O}, {K, O, O, B}, {K, O, O, K}, {K, O, O, O}, {O, B, B, B}, {O, B, B, K}, {O, B, B, O}, {O, B, K, B}, {O, B, K, K}, {O, B, K, O}, {O, B, O, B}, {O, B, O, K}, {O, B, O, O}, {O, K, B, B}, {O, K, B, K}, {O, K, B, O}, {O, K, K, B}, {O, K, K, K}, {O, K, K, O}, {O, K, O, B}, {O, K, O, K}, {O, K, O, O}, {O, O, B, B}, {O, O, B, K}, {O, O, B, O}, {O, O, K, B}, {O, O, K, K}, {O, O, K, O}, {O, O, O, B}, {O, O, O, K}, {O, O, O, O}}==81 permutation.

Apr 17, 2022
#3
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Thank you very much!

Guest Apr 17, 2022