Hi folks,

Here is a question from my A level maths text book. I can't get the right answer!

find the total number of 4 letter permutations of 4 letters selected from the word ARRANGEMENT. How many of these permutations contain 2 Rs?

Here are my calculations, can anyone spot the error?

We have 11 letters with 4 pairs (2A, 2E, 2N, 2R) and three singles G, M, T. Consider the mutually exclusive scenarios and add them up.

1) Arrangements of 4 letters in which all are different. i.e. select 4 letters from the 7 different ones.

here goes with the LaTex! hope it works.

$\displaystyle ^{7}P_{4}$ = 7!/(7 - 4)! = 7!/3! = 7 x 6 x 5 x 4 = 840 permutations.

2) Arrangements of 4 letters with one pair and two other letters

We select a pair, say the 2 Rs in one way. Then we select two letters from the available 6

i.e. $\displaystyle ^{6}P_{2} $ = 6! / 4! = 6 x 5 = 30

These 4 letters can be arranged in 4! / 2! ways = 4 x 3 = 12 ways

so we have 30 x 12 = 360 arrangements.

There are 4 pairs so the total number of 4 letter arrangements gives 360 x 4 = 1440 permutations

3) Arrangements of 4 letters containing 2 pairs

In this case we select 2 letters (each a pair) from the available 4 pairs i.e. $\displaystyle ^{4}P_{2} $

= 4! / 2! = 4 x 3 = 12 ways. These 4 letters can be arranged in 6 ways for example

AAEE, EEAA, AEAE, EAEA, AEEA, EAAE

so we have 12 x 6 = 72 permutations

So the total number of permutations should be 840 + 1440 + 72 = 2352

The number of permutations containing 2 Rs is taken directly from part 2) i.e. 360.

Both parts are wrong. The correct answers are 1596 and 198.