sanju09 wrote: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
[spoiler]Source: Eric's collection on BTG
OA A[/spoiler]
The leftmost digit is even implies it can be any digit from 2, 4, 6, 8 (that is 4 digits).
The second is odd implies it can any digit from 1, 3, 5, 7, 9 (that is 5 digits).
The third digit is a non even prime implies any digit from 3, 5, 7 (that is 3 digits).
The fourth digit can be any digit, which is not being used in the earlier 3 numbers (that is 7 digits, as 3 digits have been used before).
The fifth digit can be any digit, which is not being used in the earlier 4 numbers (that is 6 digits, as 4 digits have been used before).
Case 1: If the 2nd and 3rd digits are the same.
No. of 5 digit integers that can be formed = 4 * 3 * 1 * 8 * 7 = 672 (here 4th and 5th digits an be any digit from 8 and 7 digits respectively, as 2nd and 3rd digits are the same)
Case 2: If the 2nd and 3rd digits are not the same.
2nd digit can be any digit from 1 and 9
3rd digit can be any digit from 3, 5, and 7
No. of 5 digit integers that can be formed = 4 * 2 * 3 * 7 * 6 = 1,008
Case 3: If the 2nd and 3rd digits are from 3, 5, or 7.
2nd digit can be any digit from 3, 5, 7
3rd digit can be any digit from 3, 5, 7, but would not be the one that is already used as the 2nd digit
No. of 5 digit integers that can be formed = 4 * 3 * 2 * 7 * 6 = 1,008
Therefore, required no. of 5 digit numbers = 672 + 1,008 + 1,008 =
2,688, which is not there in the given answer choices.
