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?
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?
Thanks in advance
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.
