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12 Mar 2021, 03:16
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C
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Difficulty:
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49% (03:04) correct
51%
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wrong
based on 83
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Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the units digits of the numbers in S?
(A) 294
(B) 228
(C) 216
(D) 194
(E) 144
(A) 294
(B) 228
(C) 216
(D) 194
(E) 144
13 Mar 2021, 01:10
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Bunuel
Let the number be ABCDE.
Since we are looking at units digit, let us concentrate on position E.
1) When E is an odd number, say 1
(Odd),_,(Even),_,1 OR (Even),_(Odd),_,1
ODD: One of 3 or 5 can take any one of the position of A or C, so 2C1*2C1.
The remaining odd number can be placed at any one of the position of B or D, so 2 ways.
The way odd numbers can be placed is 2C1*2C1*2=8
EVEN: There are 2 places left for 2 even numbers after placing of odd numbers, so these 2 even numbers can be placed in 2! ways
Total numbers when 1 is in units digit = 8*2=16
Similarly, there will be 16 ways when each of 3 and 5 are the units digit.
Sum = 16(1+3+5)=16*9=144
2) When E is an even number, say 2
(Odd),_,(Odd),_,2
ODD: Choose 2 out of 3 odd numbers and arrange them in 2 ways for position A and C = 3C2*2!=6.
The remaining odd number can be placed at any one of the position of B or D, so 2 ways.
The way odd numbers can be placed is 6*2=12
EVEN: There is 1 place left for 1 even number after placing of odd numbers, so 1 way
Total numbers when 2 is in units digit = 12
Similarly, there will be 12 ways when 4 is the units digit.
Sum = 12(2+4)=12*6=72
Total: 144+72=216
C
23 Mar 2021, 05:44
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Bunuel
Solution:
Let’s first determine the number of integers with a units digit of 1. Since the units digit is an odd position, one of the following must be true: 1) the hundreds digit is odd and the ten thousands digit is even; or, 2) the ten thousands digit is odd and the hundreds digit is even.
In the first case, there are two choices for the hundreds digit (3 or 5) and two choices for the ten thousands digit (2 or 4). The tens digit can be any one of the two remaining digits and the thousands digit will be the only remaining digit. Thus, there are 2 x 2 x 2 x 1 = 8 such integers.
In the second case, there are two choices for the ten thousands (3 or 5) and two choices for the hundreds digit (2 or 4). Similar to the first case, there are also 2 x 2 x 2 x 1 = 8 such integers in this scenario.
In total, there are 8 + 8 = 16 integers with a units digit of 1 where exactly two of the odd positions are occupied by odd digits.
We can do the identical calculation as above for the case where the units digit is 3 or 5, and we will get the same result. Thus, there are 16 integers where the units digit is 3 and 16 integers where the units digit is 5.
Next, let’s determine the number of integers with a units digit of 2. Since exactly two of the odd positions must be occupied by odd digits and since one of the odd positions is occupied by an even digit, the hundreds and the ten thousands digits must contain odd integers. There are 3 choices (1, 3, or 5) for the hundreds digit and 2 choices (the two remaining odd digits) for the ten thousands digit. The tens digit can be any one of the two remaining digits, and the thousands digit will be the only remaining digit. Thus, there are 3 x 2 x 2 x 1 = 12 integers with a units digit of 2 where exactly two of the odd positions are occupied by odd digits.
Repeating the identical calculation as above for the case where the units digit is 4, there are 12 integers with a units digit of 4 as well.
Thus, the sum of the units digits in set S is:
1 x 16 + 2 x 12 + 3 x 16 + 4 x 12 + 5 x 16 = 216
Answer: C
