Monkeying around
In how many different ways can you rearrange the letters in the word MONKEY such that no letter in the new arrangement is in the same place as in the original word? (I.e. NOKEYM wouldn't work since O would be in the same place as it was in MONKEY.)
For more permutations quizzes, check out my other problems .
The answer is 265.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
5 solutions
Yes ,the same solution
Ah... Nice answer!
can also use the dearrangement formula derived using inclusion-exclusion principle ie; n!(1/2!-1/3!+1!-+-+....1/n!) here n=6 (because we need to "dearrange" 6 letters(all distinct) of the word "MONKEY"
Using the derangement formula = 6 ! ( 1 1 − 1 1 + 2 1 − 6 1 + 2 4 1 − 1 2 0 1 + 7 2 0 1 ) = 7 2 0 ( 1 4 4 5 3 ) = 2 6 5
By the derangement theorem, the number of combinations is given by !n = n!(1 - 1/1! + 1/2! - 1/3! + ....).
So, in this case, since MONKEY has six letters, the number of permutations where no letter winds up in the same place is given by:
!6 = 6!(1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!) = 265
Geoff -- nice problem! I've added it as a "Try-it-yourself" to the wiki on derangements .
Log in to reply
Excellent, thanks Eli... I hope people like it! :)
The number of derangements for n objects is e n ! , rounded to the nearest integer.
Thus, the number of derangements is e 6 ! = e 7 2 0 ≈ 2 6 5 .
Nice answer!
can also use the dearrangement formula derived using inclusion-exclusion principle ie; n!(1/2!-1/3!+1!-+-+....1/n!) here; in this question; n=6 (because we need to "dearrange" 6 letters(all distinct) of the word "MONKEY"
How would this be done for a word like "TELLER"
A formula is also D n = ( n − 1 ) ( D n − 1 + D n − 2 ) with D 0 = 1 and D 1 = 0 .