EnergySP wrote:
How many four-digit odd numbers do not use any digit more than once?
A. 1728
B. 2160
C. 2240
D. 2268
E. 2520
The answer is C, 2240.
Explanation from the book makes more sense now that I've thought about it a bit more.
Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.
8*8*7*5 = 2240
What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.
The question asks for 4 digit odd number which do not use any digit more than once.
So, the questions is basically asking to arrange the digits from 0,1,2,..,3,9 without repetition to form a 4 digit number with last digit as 1,3,5,7 or 9 only.
Lets say the 4 digit number be PQRS.
So, S can take numbers as 1,3,5,7,9 i.e. 5 ways.
Now we will chk nos. for P as it has restriction of being a non - zero. So it can take numbers as 1,2,..,9 - the no. taken in S which can be done in 8 ways,
Now, Q can take nos. 0,1,2,..., 9 minus the no. taken in S & P which can be done in 8 ways.
Now, R can take nos 0,1,2,...9 minus the no. taken in S, P & Q which can be done in 7 ways.
So, total count of such numbers = 8*8*7*5 = 2240
Answer C