1. ## Counting

Hi everyone, I have a counting problem:

Find the number of three-letter words that use letters from ten-letter set {A,B,...J} in which all letters are different and the letter appear in alphabetical order.

I know that there are 10.9.8= 720 three-letter words such that all letters are different, but I can't solve the second condition.
Can someone help me? I will be thankful.

2. ## Re: Counting

divide the set into subset of two elements like(A,B),(B,C),(C,D),(D,E),(E,F),(F,G),(G,H),(H,I )
now find the corresponding ways for which u can choose the third element
1.(A,B):8
2.(B,C):7
3.(C,D):6
4.(D,E):5
5.(E,F):4
6.(F,G):3
7.(G,H):2
8.(H,I):1

now sum all these u get 36 as no of three letter words that can be formed

3. ## Re: Counting

Originally Posted by math88
Find the number of three-letter words that use letters from ten-letter set {A,B,...J} in which all letters are different and the letter appear in alphabetical order.
Any subset of three of those ten letters can be arranged is alphabetical order.

So the answer is combination of ten choose three.

4. ## Re: Counting

but one of the combinations can be (C,A,B) they are not in alphabetical order

5. ## Re: Counting

Originally Posted by prasum
but one of the combinations can be (C,A,B) they are not in alphabetical order
If you give me a set of any three letters, I can arrange them in alphabetical order.

Consider the set $\{C,~A,~T\}$
There six ways to arrange those letters: $TAC,~TCA,~CTA,~CAT,~ATC,~\color{blue}{ACT}$
Of those six only one is in alphabetical order.

There are $(10)(9)(8)=720$ possible three letter strings. But only one in every six is in alphabetical order or $120$.

6. ## Re: Counting

Hello, math88!

Find the number of three-letter words that use letters from ten-letter set {A, B, ... J}
in which all letters are different and the letter appear in alphabetical order.

I know that there are 10.9.8 = 720 three-letter words such that all letters are different,
but I can't solve the second condition. . Sure you can!

Your list of 720 words includes 6-word sets like: {ABC, ACB, BAC, BCA, CAB, CBA}
. . Only one of them (ABC) is in alphabetical order.
Hence, your list is six times too long: .$\displaystyle 720 \div 6 \,=\,120$

You had: .$\displaystyle 10\cdot9\cdot8 \,=\,\frac{10!}{7!}$
We divide by 6: .$\displaystyle \frac{10!}{3!\,7!}$
This is the answer that Plato gave: .$\displaystyle _{10}C_3\,\text{ or }\,{10\choose3}$