number of permutations in the words IDAHO and TENNESSEE

Question

NUMBER OF PERMUTATIONS IN THE WORDS  IDAHO and TENNESSEE

Answer 1

For the word IDAHO, there are 5 letters to choose from for the first position, 4 for the second position (everything but the first letter), ..., 1 letter for the last position. That makes a total of 5*4*3*2*1 = 120 permutations (ways to order the letters). There is a special notation for this type of product, called a factorial, and labeled with an an exclamation mark, like this:

 

5! = 1*2*3*4*5

 

or, more generally,

 

n! = 1*2*3*...*(n-1)*n

 

Note that 0! = 1. This is the only choice consistent with the theorem n! = (n-1)! * n when n = 1.

 

For the word TENNESSEE, it is slightly more difficult, and there are (at least) two different ways to think about it, although the two ways I present depend on the same basic principle.

 

1) If there were 9 distinct letters, there would be 9! = 362880 ways to arrange them. So if we label each E with a number, say E1, E2, E3, and E4 and similarly with the other letters, we would have 9! permutations. But, for instance, *E1**E2E3*E4* (the *s refer to other letters) and *E2**E3E1*E4* would be the exact same string. That happens no matter where the E's are; there are 24 = 4! ways to arrange the E1, E2, E3, and E4 in the positions where the E's are, so what our calculation considers 4! separate cases is actually just one. The same collapse happens with the N's and S's, so there are really just 9!/(4!*2!*2!) permutations. Note that if you want to divide by 1! for the one T, or by 0! for a letter that is not there, you may do that, but that does not affect the answer; it is just a division by 1 if you do.

 

2) If you know about combinations, there are 9C4 = 9!/(4!*5!) ways to pick the 4 positions for the E's. Once you select the positions, there are only 5 positions left for the other letters, so there are 5C2 = 5!/(2!*3!) ways to pick the 2 positions for the N's. Similarly, there are 3C2 = 3!/(2!*1!) ways to pick the 2 positions for the S's, and the T goes in the remaining place. We get:

 

9!/(4!*5!) * 5!/(2!*3!) * 3!/(2!*1!)

 

permutations. Some factors cancel out, leaving 9!/(4!*2!*2!) permutations. The reason why this uses the same basic mechanism as way 1 is that the combination formula can be proven with the type of logic used in way 1: if you choose r objects out of n, you can assign each one of the n objects a number (1 to n), and the ones with numbers 1 through r are selected. There are n! ways to order the n objects, but r!*(n-r)! ways collapse into one because we do not care about the numbers assigned, only whether they are selected or not (so the arrangement of the numbers 1 through r is not relevant, and neither is the arrangement of the numbers r+1 through n).

 

Note that ways 1 and 2 give the same answer. This is important. If there is just one right answer, two correct ways to solve a problem should give the same answer. If they do not appear to do that, either one (or both) solutions has a mistake, or the two answers actually are equivalent.