The number of three-digit numbers with distinct digits that be formed using the digits 1,2,3,5,8 and 9 is . the probability that both the first digit and the last digit of the three-digit number are even numbers .

a)120

b)6.67%

Step-by-step explanation:

Given:

No. of digits given= 6

Digits given= 1,2,3,5,8,9

Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be

6P3= 6!/3!

      = 6*5*4*3*2*1/3*2*1

      =6*5*4

      =120

Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:

As the first and last digits can only be even

then the form of number can be

a)2n8 or

b)8n2

where n can be 1,3,5 or 9

4*2=8

so there can be 8 three-digit numbers with both the first digit and the last digit even numbers

And probability = 8/120

                          = 0.0667

                          =6.67%

The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !

1.

2.

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00