Method 1: four digit number start after 999 and end at 9999so 4 digit no. = 9999-999 = 9000method 2: - - - - there are four places in 4 digit no. In first place only 1-9 can come , for rest 0-10so total no. = 9*10*10*10=9000* * * * *Wrong!Those are the number of PERMUTATIONS, not combinations. In a combination, the order of digits does not matter. So 1234 is the same as 1342 or 4213 etc.
To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.
Let's try and figure this one out...First case (the easy one) - We want to use any of the 10 digits to create a 5-digit number. In this case, we have 5 slots and in each slot we can put any of 10 digits.So the answer is 10*10*10*10*10 = 510 = 100,000 combinationsSecond case - We want to create a 5-digit number using any of the 10 digits, but we only want to use each digit once. This turns out to be only a little more difficult.Let's look at each digit individually.First digit - We can use any digit from [0-9], so we have 10 possibilities.Second digit - We can use any digit from [0-9] except for the one used in the first digit, so we now have 9 possibilities.Third digit - We can use any digit from [0-9] except for the ones used in the first and second digits, so we now have 8 possibilities.Fourth digit - As above, except we now have 7 possible choices.Fifth digit - As above, except we now have 6 possible choices.So our final number of combinations is 10 * 9 * 8 * 7 * 6 = 30,240 combinationsSpecial case - What happens if we want to create a number which does not begin with a zero? Well, we can make a simple adjustment to either of the above cases to take care of this. Just observe that the first digit is not limited to 9 possibilities [1-9], not 10.Special first case = 9 * 10 * 10 * 10 * 10 = 90,000 combinationsSpecial second case = 9 * 9 * 8 * 7 * 6 = 27,216 combinations
There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.
0-9 = 10 numbersevery full set of 10 numbers presents a multiple of 10 variations.consider if it was a 1-digit code, using 0-9.the code could be:012345678or9that's 10 options.now if it were a 2-digit code, your 10 options are permutated 10 more times (10^2).in practise that looks like00010203...080910111213...18192021and so on.you're going from 00 to 99, so, all told, it's 100 permutations.written simply, that's: 10*10=100 or: 10^2=100now, if it's a 3-digit code, well, that's a whole notherpermutation of 10 for each of those previous 100 combinations.it looks like:000001002003...(all the way up to)...996997998999that's 000 to 999, which is 1000 unique permutations.written simply, that's: 10*10*10=1000 or better yet: 10^3=1000the logic continues for a 4-digit code, for which there would be 10,000 permutations. (10^4=10,000)if you're working with only a set of 9 numbers (for example 1-9 instead of 0-9), then you would simply apply exponents to 9 instead of 10. so a 3-digit code with numbers 1-9 has 729 possibile permutations (9^3 = 9*9*9 = 729)
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
There are 9C3 = 10*9*8/(3*2*1) = 120 of them.
Oh, dude, it's like you're asking me to do math or something. Okay, so for a 2-digit code, you have 10 options for the first digit and 10 options for the second digit, so it's like 10 times 10, which equals 100 possible combinations. But hey, who's counting, right?
it's 610, which is 60,466,176. Since each digit spot could be anything 0-9 there are 10 numbers to try for each digit, making it 610.
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PUK is Pin Unblocking Code. This is normally a code available to help mobile users with the telephone company as when one activates PIN code for his/her mobile and by chance forgets that PIN code and enters wrong PIN code for 4 times then PUK code is asked. This code can be given to the user on request by the customer care after proper verification. PUK code is 10 digit code and if user enters wrong PUK code for 10 times then SIM card is bloked and user needs to change the SIM card. Asmita
PUK is Pin Unblocking Code. This is normally a code available to help mobile users with the telephone company as when one activates PIN code for his/her mobile and by chance forgets that PIN code and enters wrong PIN code for 4 times then PUK code is asked. This code can be given to the user on request by the customer care after proper verification. PUK code is 10 digit code and if user enters wrong PUK code for 10 times then SIM card is bloked and user needs to change the SIM card. Asmita
There are 10 possible digits that can be used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. If you can only use each digit once, there are 10 choices for the first digit, 9 remaining choices for the second digit, and 8 remaining choices for the final digit - 10 x 9 x 8 = 720 combinations. If you can use each digit repeatedly, you have 10 choices for each digit - 10 x 10 x 10 = 1000 combinations.
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