Question

A three-digit number is formed from nine numbers (1, 2, 3, 4, 5, 6, 7, 8 & 9). No number can be repeated. How many different three-digit numbers are possible if 1 and 2 will not be chosen together?

Select one:

a. 462

b. 336

c. 672

d. 210

Answer 1

solution:

From the given information

Total No.of possible numbers = 9

No.of digits in required number = 3

step-1 : Let us calculate the total no.of 3 digit numbers possible from 9 numbers  

1st Digit

2nd Digit

3rd Digit

----> No.of possible numbers at 1st digit = 9

----> No.of possible numbers at 2nd digit = 8 , since already one number assigned at 1st digit

----> No.of possible numbers at 3rd digit = 7 , since already 2 numbers assigned at 1st and 2nd digits

----> By the product rule

Total no.of 3 digit numbers possible from 9 numbers = 9 * 8 * 7 = 504

step-2 : Let's calculate No.of digits possible if 1 and 2 choosen together

we can think of 6 possible conditions, where we can choose 1 and 2 together

1

2

1

2

1

2

2

1

2

1

2

1

At each empty box

No.of possible digits = 9 - 2 = 7

Total No.of digits possible if 1 and 2 choosen together = 6*7 = 42

step-3:

Total No.of 3 digit numbers possible if 1 and 2 will not choosen together = 504 - 42 = 462

  Option-A is correct