
The number 2006 is made up of exactly two zeroes and two other digits whose sum is 8. The number of 4 digit numbers with these properties (including 2006) is:
a.7
b.18
c.21
d.24
Answer
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Hint: In this question consider the fact that the four digit number cannot start with 0, as then in that case it won’t be a 4 digit number thus zero can come in the middle or in the last of the number. Now for the sum to be 8 the digits are gathered as 17, 26, 35, 44, 53, 62, and 71. Thus consider the possible cases zero is at second place and other zero is at fourth place, then the case in which one zero is at second place and other zero is at third place and the last case in which one zero is at third place and other zero is at fourth place. This will help approaching the problem.
Complete step-by-step answer:
We have to calculate all these numbers including (2006) which is a given number in this format.
As we know numbers cannot start from zero otherwise it becomes a two or three digit number which is not our criteria.
So zero can come in the middle or in the last.
And the digits whose sum is 8 can we written as (17, 26, 35, 44, 53, 62, and 71)
So there are 7 possibilities that the sum of the two digits is eight not including the possibilities of (08 and 80) otherwise there will be three zeroes in the 4 digit number which is not our desired case.
Case – 1: When one zero is at second place and other zero is at fourth place so the number of four digit numbers are:
1070, 2060, 3050, 4040, 5030, 6020, and 7010
So there are 7 possibilities in case 1.
Case – 2: When one zero is at second place and other zero is at third place so the number of four digit numbers are:
1007, 2006, 3005, 4004, 5003, 6002 and 7001.
So there are another 7 possibilities in case 2.
Case – 3: When one zero is at third place and other zero is at fourth place so the number of four digit numbers are:
1700, 2600, 3500, 4400, 5300, 6200 and 7100.
So there are another 7 possibilities in case 3.
So the total 4 digit numbers with the given criteria = 7 + 7 + 7 = 21.
So this is the required answer.
Hence option (C) is the correct answer.
Note:Since in this question the pattern that has to be followed was of 2006, thus it was mandatory to take up two zeroes while the rest numbers whose sum is 8 are only taken into consideration as discussed above that is 17, 26, 35, 44, 53, 62, and 71.
Complete step-by-step answer:
We have to calculate all these numbers including (2006) which is a given number in this format.
As we know numbers cannot start from zero otherwise it becomes a two or three digit number which is not our criteria.
So zero can come in the middle or in the last.
And the digits whose sum is 8 can we written as (17, 26, 35, 44, 53, 62, and 71)
So there are 7 possibilities that the sum of the two digits is eight not including the possibilities of (08 and 80) otherwise there will be three zeroes in the 4 digit number which is not our desired case.
Case – 1: When one zero is at second place and other zero is at fourth place so the number of four digit numbers are:
1070, 2060, 3050, 4040, 5030, 6020, and 7010
So there are 7 possibilities in case 1.
Case – 2: When one zero is at second place and other zero is at third place so the number of four digit numbers are:
1007, 2006, 3005, 4004, 5003, 6002 and 7001.
So there are another 7 possibilities in case 2.
Case – 3: When one zero is at third place and other zero is at fourth place so the number of four digit numbers are:
1700, 2600, 3500, 4400, 5300, 6200 and 7100.
So there are another 7 possibilities in case 3.
So the total 4 digit numbers with the given criteria = 7 + 7 + 7 = 21.
So this is the required answer.
Hence option (C) is the correct answer.
Note:Since in this question the pattern that has to be followed was of 2006, thus it was mandatory to take up two zeroes while the rest numbers whose sum is 8 are only taken into consideration as discussed above that is 17, 26, 35, 44, 53, 62, and 71.
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