**Suggested Approach and Solution:-**

For questions with special conditions, a good approach is to consider to the special conditions first. It matters whether the last digit is a ‘3’ or not. This is a complication that is handled by a Case-by-case Analysis and then totaling up the number of possibilities.

__Case 1__: 3 is the last digit

If 3 is the last digit, we have 1 choice (i.e. Hobson’s choice) for the last digit. The first digit must be either a 1 or a 5 i.e. 2 choices. The remaining 5 digits can be filled in

^{5}P_{5}= 5!= 5 x 4 x 3 x 2 x 1 = 120 ways.

Number of ways for this case = 2 x 5! = 240 ways.

__Case 2__: 3 is not the last digit

In this case, the last digit must be either a 1 or a 5 i.e. 2 choices. Because these digits only occur once, the first digit will definitely be different from the last digit. So we need not worry about the first-digit condition as it is automatically satisfied. The front 6 digits can be filled in

^{6!}/

_{2!}= 360

We divide by 2! because the digit 3 is definitely repeated.

Number of ways for this case = 2 x 360 = 720 ways.

The total number of ways = 240 + 720 = 960.

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