The sum you require is 4!/2!2!, which is 6. This means there are 6 different two-digit numbers.
* * * * *
Wrong! The above formula is for the number of combinations, not permutations. However, since the number 23 is different from the number 32, you require permutations and not combinations.
There are 4*3 = 12
23, 25, 27,
32, 35, 37,
52, 53, 57,
72, 73, 75.
24
27 three digit numbers from the digits 3, 5, 7 including repetitions.
24
128
2,401
There are 4*3 = 12 such numbers.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
99999
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
500
24
Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits. There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively. So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120 Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360. the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720. The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720. So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.
840
There are 60480 numbers.
1 set
It is 120 if the digits cannot be repeated.
Assuming the question requires the biggest number formed using the given digits - rather than applying mathematical operations to them - the answer is 76430.