My Math Forum Question on permutation and combination

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 15th, 2010, 07:53 AM #1 Newbie   Joined: May 2010 Posts: 11 Thanks: 0 Question on permutation and combination Hi, all I am not being able to solve the following problem, please help. 10 different letters of an alphabet are given. Words with 5 letters are formed from these given letters. Then what will be the number of words which have at least 1 letter repetition?
May 15th, 2010, 08:51 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Question on permutation and combination

Hello, rosalie!

Quote:
 10 different letters of an alphabet are given. 5-letter words are formed from these given letters. What is the number of words which have at least 1 letter repeated ?

For each of the five letters, there are 10 choices of letters.
[color=beige]. . [/color]$\text{There are: }\:10^5 \:=\:100,000\text{ possible 5-letter words.}$

$\text{How many of these words have }no\text{ letters repeated?}$

$\text{There are: }\:10\cdot9\cdot8\cdot7\cdot6 \:=\:30,240 \text{ words with }no \text{ repeated letters.}$

$\text{Therefore, there are: }\:100,000\,-\,30,240 \:=\:69,760\text{ words with }some\text{ repeated letters.}$

 May 15th, 2010, 06:09 PM #3 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Re: Question on permutation and combination Hi, Could you explain to me why 10^5 is the total amount of 5 letter words ? wulfgarpro.
 May 15th, 2010, 08:04 PM #4 Newbie   Joined: May 2010 Posts: 11 Thanks: 0 Re: Question on permutation and combination Hi, soroban Thanks for giving me the solution, but I didn't get your point.Could you please explain me in details.I m new 2 permutation and combination so I need explanation.Why did write 10^5 and 10.9.8.7.6.Why r these so?Please help me. Thanks again.
 May 15th, 2010, 09:48 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Question on permutation and combination Given that the same letter may be chosen more than once you have ten choices for the first letter, ten choices for the second letter and so on, up to five letters (10^5). To get the amount of words that do NOT have repeated letters you have ten choices for the first letter, nine choices for the second letter, eight choices for the third letter and so on (10 * 9 * 8 * 7 * 6).
 May 15th, 2010, 10:23 PM #6 Newbie   Joined: May 2010 Posts: 11 Thanks: 0 Re: Question on permutation and combination Hi, greg1313 Thanks ,now I got 10^2 , but didn't get 10*9*8*7*6.Could u please explain it in detail?Please help. Thanks again.
May 15th, 2010, 10:32 PM   #7
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,948
Thanks: 1139

Math Focus: Elementary mathematics and beyond
Re: Question on permutation and combination

Quote:
 Originally Posted by rosalie Hi, greg1313 Thanks ,now I got 10^2
Don't you mean 10^5?

Quote:
 Originally Posted by rosalie . . . didn't get 10*9*8*7*6.Could u please explain it in detail?
For example, let's use the five letters M A T H S. These letters are all different (of course!). So, if you pick M, then there are only four letters left to choose from, since we don't want repeating letters. Got it?

 May 16th, 2010, 10:52 PM #8 Newbie   Joined: May 2010 Posts: 11 Thanks: 0 Re: Question on permutation and combination Hi, greg1313 Thanks a lot for giving me details of my queries.Now , I got it.Sorry, I wrote 10^2 instead of 10^5 in my previous post.It was just a typing mistake. Thanks again.

permutation&combination alphabet

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post happy21 Algebra 2 November 12th, 2013 09:17 AM b_andries Algebra 8 April 19th, 2011 06:36 AM daivinhtran Algebra 1 April 10th, 2011 01:46 PM mikeportnoy Advanced Statistics 1 September 27th, 2008 11:45 AM binshehab Abstract Algebra 2 June 5th, 2008 02:29 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top