Given the individual digits: 4, 6, 9, 1, and 3,
the number of 4-digit permutations is (5! / 1!) = (5 x 4 x 3 x 2) = 120 .
The number of combinations is [ 5! / (1!4!) ] = 5 .
"Permutations" do consider the order. If you want the order ignored, then you're looking for "combinations". 3-digit permutations, repetition allowed: 9 x 9 x 9 = 729 3-digit permutations, no repetition: 9 x 8 x 7 = 504 3-digit combinations, no repetition = 504/6 = 84.
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
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.
To calculate the number of different 4-digit combinations that can be made using numbers 0 through 9, we use the concept of permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the number of options for each digit (10 in this case) and r is the number of digits (4 in this case). Therefore, the number of different 4-digit combinations that can be made using numbers 0 through 9 is 10^4, which equals 10,000 combinations.
9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.
10! permutations of the word "Arithmetic" may be made.
There are 7893600 permutations.
There are 5,273,912,160 permutations of 5 numbers out of 90.
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There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
2520
All the numbers from 0 to 999 so 1000 numbers However if you mean only using each number once the answer is about 720 * * * * * No. The above answer refers to the number of permutations. Permutations are NOT the same as combinations as anyone who has studies any probability theory can tell you. The number of combinations is 9C3 = 9*8*7/3*2*1 = 84