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.
There are 5*4*3 = 60 such numbers.
24
Assuming each of the given digits can be used only once, the answer is 24. If not, the answer is infinity.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
24
There are 4*3 = 12 such numbers.
99999
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
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.
There are 5*4*3 = 60 such numbers.
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
24
500
There are 60480 numbers.
840
Assuming the question requires the biggest number formed using the given digits - rather than applying mathematical operations to them - the answer is 76430.
1 set