Answer to Question #176752 in Statistics and Probability for Hiii

Question

1 a) i) Find how many numbers there are between 100 and 999 in which all three digits are different.

ii) Find how many of the numbers in part (i) are odd numbers greater than 700.

Answer 1

(i) Numbers must be between 100 and 999, i.e. 3 digit numbers.

Let three digit number be ‘ABC’.

‘A’ cannot be zero, so it can take values from 1 to 9. There are 9 possibilities.

‘B’ can be any digit except ‘A’. There are 9 possibilities.

‘C’ can be any digit except ‘A’ and ‘B’. There are 8 possibilities.

Total number of outcomes = 9 × 9 × 8 = 648.

Answer: there are 648 numbers between 100 and 999 in which all three digits are different.

(ii) Numbers must be greater than 700, i.e. 3 digit numbers with the first number 7, 8 or 9.

Let three digit number be ‘ABC’.

1) First number is 7 or 9.

’A’ can be 7 or 9. There are 2 possibilities.

‘C’ can be any odd digit except ‘A’. There are 4 possibilities.

’B’ can be any digit except ‘A’ and ‘C’. There are 8 possibilities.

Total number of outcomes = 2 × 8 × 4 = 64.

2) First number is 8.

’A’ can be 8. There are 1 possibility.

‘C’ can be any odd digit. There are 5 possibilities.

’B’ can be any digit except ‘A’ and ‘C’. There are 8 possibilities.

Total number of outcomes = 1 × 8 × 5 = 40.

There are 64+40=104 numbers.

Answer: there are 104 odd numbers between 700 and 999 in which all three digits are different.