That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.
There are 504 of them.
yes.
expanded form
6 possible 3 digit combonations
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.
There are 504 of them.
yes.
expanded form
90 of them.
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
6 possible 3 digit combonations
5
dominant
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
Generally, we use expanded form when numbers have more than one digit to show how they're constructed. 0.6 has only one digit, so expanded form doesn't really apply.