How many numbers can you form using the digits 1, 5, 6 and 9 without repetition of digits such that the value of the digit 6 in all those number is 100 times the value of the digit 6 in the number 9561?

6 times

Step-by-step explanation:

The value of 6 in the final number should be 100 times the value of 6 in 9561

in 9561 , the value of 6 is 60

100 times 60 is 6000

Hence, the value of 6 in the final number should be 6000

Now, we have the value of one digit out of 4 for our final number

So, our number will look like: 6 _ _ _

these dashes need to be filled by the remaining numbers: 1, 5 , 9

So, we will permute 1 , 5 and 9 in these 3 places since the position of 6 is permanent

3P3 = 3! = 6

Therefore, there are 6 possible 4-digit numbers which have the value of 6 as 6000 and no number are repeated from the given set of (1, 5, 6 , 9)

x = -2

y = 848

step-by-step explanation:

plug in the numbers

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

it alot will look like this 4.4