Topic review (newest first)
- 2006-01-16 01:00:51
That's my biggest pet peeve ever. Reading a question (especially math!) wrong. I absolutely hate when I do it.
- 2006-01-15 22:11:17
Ah, reading again the M.ANAND KUMAR post I understand that he wanted to use any digit in his 8 digits phone number. Sorry, I was wrong, I thought that the 8 digits are constant. Your answer is correct.
- 2006-01-15 22:05:06
I checked again and it is correct - the formula is n!/k! where n is the number of all digits and k is the number of the repeated one. And here we are talking about exactly 8 digits that are used: for example we have the number
98347836 - and we want to keep the first 2 digits constants and we can permute the other ones: 3,4,7,8,3,6.
and here we have 2 threes, so the formula is 6!/(2*1).
You can count them to see if I am wrong.
- 2006-01-15 08:44:19
I don't think that's right.
Let's start with one digit. You can go from 0 to 9, so you have 10 choices.
Now two digits. Each are 0-9, with which you can represent all numbers from 00 to 99, 100 choices.
Same for three digits. 1000 choices.
And so on. The answer is 10^x, where x is the number of digits. 10^6 in this case.
- 2006-01-15 06:42:55
OK, if you have 8 digits number and you want to keep the first 2 digits constants here is what you do:
You have left 6 digits to permute. If the 6 left digits are unique your formula is n! where n equals 6. So you have 6*5*4*3*2*1 different 8 digits numbers. BUT when you have reapeated digits and their number is k, your formula is n!/k!
I hope it makes sence.
- M.ANAND KUMAR
- 2006-01-15 05:25:41
I Did Not Have Clarity In Permutation And Combination, Although I Would Get The Results Correct At Many Times, But Your Explanation
Of Watering 7 X 4 Combination Made It Simpler And Easier.
I Would Like To Ask You To Explain Me One More Puzzle.
There Are Eight Digits In Telephone Numbers.
Keeping The First Two Digit Constant, How Many Different Telephone 8 Digit Numbers Can Be Formed. What Is The Formulae.
Expecting Your Reply
- 2005-09-28 23:45:54
You just need to be careful if you're given a word with repeated letters, like 'deeded'.
- 2005-08-29 22:42:37
thank you so much, thats really helpful!! i will definately be using this sight again if i have a conundrum!! A+++++!!!
- 2005-08-29 21:39:18
This is covered under the subject of "Permutations" ... a Permutation is like an ordinary combination but where the order matters
If I have understood your question correctly, you are asking how many ways can you arrange (without repeats) 4 letters chosen from the 7 letters "ATERING" (because you are told you have to choose the "W" and place it at the end)
So let us ignore the "W", then, and think about how many permutations of 4 you could have in 7
Now, think about this: how many combinations of 1 could be found in 7? Just 7 (A,T,E,R,I,N,G)
and how how many combinations of 2 could be found? 7×6 = 42 (because after having chosen one letter there are only 6 left to choose from)
and how how many combinations of 3 could be found? 7×6×5 = 210
and how how many combinations of 4 could be found? 7×6×5×4 = 840
And that is your answer
BTW There is a formula for the total number of permutations: P(n, r) = n! / (n-r)! , where "!" means factorial (which is calculated like this 4!=4×3×2×1)
In your case n=7 and r=4, so P = 7!/(7-4)! = 7!/3! = (7×6×5×4×3×2×1) / (3×2×1) = 7×6×5×4 = 840
And of course, you could always just list all possibilities. Let me see: ATERW, ATREW, ARTEW, RETAW ... *fades into distance*
- 2005-08-29 20:29:53
hi there, i have a problem, i cant answer this question to save myself!
5 letters are chosen from the word WATERING and placed in a row, the number of ways in which this can be done if the last letter is to be W is:
thankyou very much for your help!