Question

Question, doubt

Answer 1

the red part

Answer 2

please explain.

Answer 3

@wolf1728 @ParthKohli @phi

Answer 4

@ganeshie8

Answer 5

ya tell me

Answer 6

When several elements are alike and we attempt to rearrange them , the resulting permutation does not change a bit. Say, we have ,

\(A\) \(A\) \(B\) \(E\) \(C\) \(A\) \(D\)

\(7\) letters, but \(3\) of them are same ( \(A\) ) .

Note that, when we swap \(2\) \(A's\) , there's no change at all. Thus, for each of the \(3!\) permutations of the \(3\) \(A\;s\) , we'd get the same arrangement. Thus, which was previously \(7!\) is now \(\frac{7!}{3!}\) .

It would be easier to visualize with this example -

\(A\) \(A\) \(B\) \(.....(1)\)

Consider this parallel to -

\(A\) \(B\) \(C\)


We'd permute \( (1) \) -

\(A\) \(A\) \(B\)

\(A\) \(B\) \(A\)

\(B\) \(A\) \(A\)

Note that, in each case, we could have swapped the \(A \; s\) ( \(2!\) ), which we would have done in case of \(A\;\;\;B\;\;\;C\) .

Thus, we have to divide \(3!\) by \(2!\) , since there are \(2!\) abundant cases opposite to each cases.

Answer 7

wow, nice @Shouborno

Answer 8

Just check if my concept is going right way or not

Answer 9

If A B B C A A B
then

Answer 10

its \[\frac{7!}{3!3! }\]

Answer 11

@Shouborno @Shouborno

Answer 12

@Qwertty123 @Wolfman77 @Wolfboy

Answer 13

@wolf1728

Answer 14

@hartnn
Please help me in this, am I going in the right direction?

Answer 15

yes, thats correct :)

MISSISSIPPI is a classic example!