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).

98 347836

98 427836

98 428736

....

You can count them to see if I am wrong.

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.

]]>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.

]]>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

Regards

Mak

Bahrain

lioness]]>

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*

]]>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:

a) 840

b) 2520

c) 1008

d) 40

e) 625

thankyou very much for your help!

lioness

]]>