A four-digit number is to be formed with the digits {0, 1, 2, 3, 4, 5} without repetition. What is the probability that this number will leave a remainder 2 when divided by 9?
Answer: 0.14
Explanation
Required probability = (number of four digit numbers that leave a remainder 2 when divided by 9) / (total number of four digit numbers that can be formed)
If a four-digit number formed with the digits {0, 1, 2, 3, 4, 5} leaves a remainder 2 when divided by 9 then the sum of the digits of the number should be 11.
This is possible only when the digits are (0, 2, 4, 5) or (1, 2, 3, 5).
This is possible only when the digits are (0, 2, 4, 5) or (1, 2, 3, 5).
With (0, 2, 4, 5) as the digits the number of four digit numbers that can be formed is 3 * 3! = 18
With (1, 2, 3, 5) as the digits the number of four digit numbers that can be formed is 4! = 24
Hence, the total number of four digit numbers satisfying the condition that can be formed is 18 + 24 = 42.
Now, the total number of four digit numbers that can be formed with the digits {0, 1, 2, 3, 4, 5} is 5*5*4*3 = 300.
Hence, the required probability = 42/300 = 0.14.
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Thanks for the post but as i am novice in Math, I would like to know how the (4! = 24) equation works.