permutation question

**hello** · June 1st 2009, 01:45 PM

if you have the letters A,B,C,D,E and you form 3 letter "words" with these letters, what is the probability that the word starts with D?

how do I show this using permutations?

using fractions I know the answer is $\frac{1}{5}$ (1/5 *4/4*3/3)

so it would be $\frac{something}{_{5}P_{3}}$

how do I show the number of possible combinations that begin with D as a permutation?

**Plato** · June 1st 2009, 02:27 PM

Originally Posted by hello

if you have the letters A,B,C,D,E and you form 3 letter "words" with these letters, what is the probability that the word starts with D?

I assume that you mean three different letters.
There are $(5)(4)(3)=60$ possible three letter words.
There are $D(4)(3)=12$ that begin with D.

**hello** · June 1st 2009, 02:52 PM

Originally Posted by Plato

I assume that you mean three different letters.
There are $(5)(4)(3)=60$ possible three letter words.
There are $D(4)(3)=12$ that begin with D.

thus the probability of making a word that begins with D is $\frac{12}{60}$

how do I show this using permutations, though?

$\frac{possibilities with D as first letter}{_{5}P_{3}}$

I know the numerator is 12 since $_{5}P_{3}$ = 60

I just need to use another permutation for the numerator. Sorry if my OP was confusing

**Plato** · June 1st 2009, 02:56 PM

Originally Posted by hello

thus the probability of making a word that begins with D is $\frac{12}{60}$ YES
I just need to use another permutation for the numerator.

Because D has be used, you have $_4 P_2 =12$

**hello** · June 1st 2009, 04:16 PM

Ah ok. Thanks

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