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Integers.

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Integers.

by neya » Wed Aug 03, 2011 3:10 am
How many integers greater than 999 but not greater than 4000 can be formed with the digits 0,1,2,3 and 4 if repetition of digits is allowed?
1.499
2.500
3.375
4.376
5.501
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by edge » Wed Aug 03, 2011 3:53 am
We have to find the number of 4-digit integers between 1000 and 3999 (inclusive) such that the digits are 0, 1, 2, 3 or 4 (with repetition allowed).

For the 1000 range:
first digit can be chosen in 1 way (can be only 1)
second digit can be chosen in 5 ways (0, 1, 2, 3 or 4)
third digit can be chosen in 5 ways
fourth digit can be chosen in 5 days

Number of possibilities = 125. Similarly, for 2000 and 3000 ranges, number of possibilities = 125.

Hence, total possibilities = 125*3 = 375; the answer is C.

EDIT: I always make this mistake of not comprehending the question completely. 4000 is included, hence the answer is 376 as Anurag posted below.
Last edited by edge on Wed Aug 03, 2011 4:14 am, edited 1 time in total.
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by Anurag@Gurome » Wed Aug 03, 2011 3:55 am
neya wrote:How many integers greater than 999 but not greater than 4000 can be formed with the digits 0,1,2,3 and 4 if repetition of digits is allowed?
1.499
2.500
3.375
4.376
5.501
The smallest integer is 1000 and the largest integer is 4000, both of which are 4 digit integers.
Except from the integer 4000, the thousands place in the 4-digit integer can be any digit from 1, 2, or 3.
Hundreds, tens and units place can be any digit from the 5 digits: 0, 1, 2, 3, and 4
Therefore, no. of 4 digit integers from 1000 to 3999 = 3 * 5 * 5 * 5 = 375
Adding 4000 to these 375 digits, we get the required total no. of integers = 375 + 1 = 376

The correct answer is 4.
Anurag Mairal, Ph.D., MBA
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