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13 Aug 2007, 06:10
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
62% (02:57) correct
38%
(01:19)
wrong
based on 350
sessions
History
Date
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Not Attempted Yet
How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?
a) 2520
b) 3150
c) 3360
d) 6000
e) 7500
a) 2520
b) 3150
c) 3360
d) 6000
e) 7500
In my way:C(4,1)C(5,1)C(3,1)A(7,2)+C(4,1)C(5,1)C(3,1)A(8,2)
But there is no correct answer in my way...
But there is no correct answer in my way...
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25 May 2010, 04:07
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RACHITKH
2688 is the correct answer, but it's not in answer choices, so OA for this question is wrong.
This question was posted in another topic (permutations-combinations-probability-download-questions-57156-20.html), below is my solution from there:
How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?
Note that the second and the third digits can be the same (in 3 cases out of 15).
1. second and third digits are the same:
4*3*1*8*7
2. second and third digits are not the same:
4*4*3*7*6
(Think about it this way: we don't want second and third digits to be the same, let's first choose the third digit (clearly it doesn't matter which one we choose first) how many possibilities are there? 3 (3 odd primes), than choose the second digit how many possibilities are there as one odd prime is already used? 5-1=4 --> 3*4=4*3.)
Total=4*3*1*8*7+4*4*3*7*6=2688.
General Discussion
24 May 2010, 22:44
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I got 2688..
Following is my calculation:
First digit can be 2,4,6,8
2nd digit can be 1,3,5,7,9
3rd digit can be 3,5,7
Notice that 2nd and 3rd digit have same numbers thus there is a possibilty of repeat numbers. Only 1 and 9 are not repeated.
Thus 4*2*3*7*6 =1008 (This considers the case when 2nd digit is 1 or 9)
Now consider the second case where the second digit is 3, 5 or 7. In this scenario 2nd and 3rd digit can be same.
Thus we get 2 equations, one when the number repeats and the other when the number doesn't repeat
4*3*2*7*6 =1008 (2nd digit is 3,5,7 and 3rd digit is different than 2nd)
4*3*1*8*7 =672 (2nd digit is 3,5,7 and 3rd digit is same as 2nd)
Adding 1008+1008+672= 2688
Hence the Answer
Following is my calculation:
First digit can be 2,4,6,8
2nd digit can be 1,3,5,7,9
3rd digit can be 3,5,7
Notice that 2nd and 3rd digit have same numbers thus there is a possibilty of repeat numbers. Only 1 and 9 are not repeated.
Thus 4*2*3*7*6 =1008 (This considers the case when 2nd digit is 1 or 9)
Now consider the second case where the second digit is 3, 5 or 7. In this scenario 2nd and 3rd digit can be same.
Thus we get 2 equations, one when the number repeats and the other when the number doesn't repeat
4*3*2*7*6 =1008 (2nd digit is 3,5,7 and 3rd digit is different than 2nd)
4*3*1*8*7 =672 (2nd digit is 3,5,7 and 3rd digit is same as 2nd)
Adding 1008+1008+672= 2688
Hence the Answer
