How many Distinct Ways are there to rearrange ALL the letters in ... (Permutations with Repetition)

How many ways are there to rearrange all of the letters in SWITCH?
in CANADA? in BANANA? in MATHEMATICS? in TORONTO?
* Take the factorial of the number of letters in the word
* DIVIDE by the factorial of all doubles, triples etc. individually.

[@yashasvasharma4689]

This was incredibly helpful thank you, Must watch!!!

[@badtzm]

this was so incredibly helpful thank you!!

[@mariyogesh827]

Thank you so much

[@cheetodust8576]

Thank you for this!

[@aadityakiran_s]

Hey, so how do you find out the total possible ways to arrange a 3 letter word form the letters of the word CANADA?

[@mroldridge]

You’re going to have to break that into cases.
Case 1: No A’s (3! = 6)
Case 2: One A (3C2 x 3! = 18)
Case 3: Two A’s (3C1 x 3!/2! = 9)
Case 4: Three A’s (3!/3! = 1)
Then add all those up, 6+18+9+1= 34

[@aadityakiran_s]

Oh, so if there were multiple repeated letters, then this would become quite complicated huh? I was trying to solve this programmatically, I suppose there's no general formula. Thanks for the info. @@mroldridge

[@huyanhle]

@@aadityakiran_s
CANADA = 6 letters. A = 3 letters . CND = 3 letters
Case 1: NO "A" at all in 3 letter word ==> CND = 3 letters to be put into 3 places and no repetition of letters ==> 3 x 2 x 1 = 3! = 6
Case 2: One "A" in 3 letter word ==> ACND = 4 letters (Note: A = 3 letters) ==> 3 (for 3 "A") x 3 x 2 = 18
Case 3: Two "A" in 3 letter word ==> AACND = 5 letters (Note: A = 3 letters). Also note repetition of 2 "A" need to be cancelled out by 2! ==> 3 (for 3 "A") x 3 x 2 / 2! = 9
Case 4: Three "A" in 3 letter word ==> AAACND = 6 letters (Note: A = 3 letters). Also note repetition of 3 "A" need to be cancelled out by 3! ==> 3 x 2 x 1 / 3! = 1
==> Answer = 6 + 18 + 9 + 1= 34

[@GodSahil]

b-but how does the multiplication principle work?

[@loveelyy8348]

be more specific pls

[@GodSahil]

@@loveelyy8348 is there any derivation of this