If the permutations of the letters in the word sure are numbered 1 through 24 in alphabetical order, what number is ruse. the answer is number 12. explain!

Every possible combination of the letters SURE are going to be listed in alphabetical order. The permutation we want is RUSE which begins with the letter R and will come after every permutation that begins with E since it is the next alphabetically. We can first determine how many permutations begin with E.

Since we start with E, there are only three letters left to form the rest of the permutation. So 3! = 3*2*1 = 6 states that there are 6 permutations that can be made from the remaining three letters. So there will be 6 permutations  that begin with E.

Using this same logic, we now know that there are 6 permutations that begin with the letter R. The letters USE are in reverse alphabetical order, which means that the word RUSE will appear as the last permutation that begins with R.

We know there are 6 permutations that begin with E, followed by 6 permutations that begin with R, making 12 total at this point. And since RUSE appears as the last permutation beginning we R, we know that RUSE shows up 12th.

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .