Abank issues 10-digit account numbers to its customers. each number is an even number, which starts with 9. repetition of digits is not allowed. how many such numbers are possible? plz

Here we use the counting principle. We consider the number of possibilities for each digit and multiply them.

Let’s think about the first digit. We are told it must be 9 so there is only 1 choice for the first digit.

Let’s consider the last digit. Since the number is even the last digit must be even so it must be 0,2,4,6 or 8. That’s 5 choices.

Now there are 10 numbers we can use : 0,1,2,3,4,5,6,7,8,9 and we have used 2 of them. That leaves 8 and since the digits can’t repeat there are 8 ways to pick the second digit, and 7 ways to pick the third...keep this up and we have 6 ways to pick the fourth, 5 ways to pick the fifth, 4 ways to pick the sixth and so on. Remember there’s five ways to pick the last digit.

That means there are: (1)(8)(7)(6)(5)(4)(3)(2)(1)(5) possible account numbers. Multiplying yields: 201,600 possible account numbers.

A) for the n row: 22+(n-1)×6therefore,b) 100-22=78that means (n-1)×6=7878÷6=1313+1=1414th row has 100 seats