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07 Aug 2024, 15:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (03:02) correct
59%
(02:51)
wrong
based on 73
sessions
History
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Arun wants to form a code using the numbers 1, 2, 3, 4, 5, and 6 as the digits of the code. He wants his code to have 5 digits. Further, he does not want to use the same number twice in his code and he does not want his code to be an odd multiple of 3. The number of codes that Arun can form is
A. 360
B. 420
C. 480
D. 540
E. 600
A. 360
B. 420
C. 480
D. 540
E. 600
08 Aug 2024, 04:28
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The easiest approach to this question is to solve the total possible combinations of 5 numbers from 6, and then to subtract all possible combinations of odd numbers which are divisible by 3.
Total Combinations of 5 numbers from 6:
Selecting 5 numbers from a possible 6 numbers: \(\frac{6!}{5!1!} = 6\).
However, this is the number of unordered combinations. As order matters in this instance, one needs to multiply through by 5!:
\(6*5! = 6*120 = 720\)
Total Combinations which are Odd and Divisible by 3:
There are three odd numbers in the set, 1; 3 & 5. These values will have to be the fifth number in the code for it to be odd. Additionally, a number is divisible by 3 when its digits sum to a value which is divisible by 3.
To find the five numbers which when summed are divisble by 3, sum the smallest five and the largest five to establish a range:
\(1+2+3+4+5 = 15\) and \(2+3+4+5+6 = 20\). Which means the only two sums which are divisible by 3 will be 15 and 18.
15 is already covered in \(1+2+3+4+5 = 15\)
To find 18, the set sums to \(1+2+3+4+5+6 = 21\). From 21 to get to 18 one simply removes 3. So \(1+2+4+5+6 = 18\).
From here one creates two subsets:
1) \(1, 2, 3, 4, 5 \)
2) \(1, 2, 4, 5, 6 \)
In 1) one has all 3 of the odd numbers which means that there will be a total of \(3*4!\) possibilities. This is because each of the three odd numbers will fill the fifth position, which leaves 4 numbers for which all possible combinations need to be accounted for.
In 2) there are two odd numbers, which means that there will be a total of \(2*4!\) possibilities.
Total combinations which are odd values divisible by 3: \(3*4!+2*4! = 5*4! = 120\)
The number of codes that Arun can form:
\(720 - 120 = 600\)
ANSWER E
Total Combinations of 5 numbers from 6:
Selecting 5 numbers from a possible 6 numbers: \(\frac{6!}{5!1!} = 6\).
However, this is the number of unordered combinations. As order matters in this instance, one needs to multiply through by 5!:
\(6*5! = 6*120 = 720\)
Total Combinations which are Odd and Divisible by 3:
There are three odd numbers in the set, 1; 3 & 5. These values will have to be the fifth number in the code for it to be odd. Additionally, a number is divisible by 3 when its digits sum to a value which is divisible by 3.
To find the five numbers which when summed are divisble by 3, sum the smallest five and the largest five to establish a range:
\(1+2+3+4+5 = 15\) and \(2+3+4+5+6 = 20\). Which means the only two sums which are divisible by 3 will be 15 and 18.
15 is already covered in \(1+2+3+4+5 = 15\)
To find 18, the set sums to \(1+2+3+4+5+6 = 21\). From 21 to get to 18 one simply removes 3. So \(1+2+4+5+6 = 18\).
From here one creates two subsets:
1) \(1, 2, 3, 4, 5 \)
2) \(1, 2, 4, 5, 6 \)
In 1) one has all 3 of the odd numbers which means that there will be a total of \(3*4!\) possibilities. This is because each of the three odd numbers will fill the fifth position, which leaves 4 numbers for which all possible combinations need to be accounted for.
In 2) there are two odd numbers, which means that there will be a total of \(2*4!\) possibilities.
Total combinations which are odd values divisible by 3: \(3*4!+2*4! = 5*4! = 120\)
The number of codes that Arun can form:
\(720 - 120 = 600\)
ANSWER E
08 Sep 2024, 10:38
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We can make 6 sets of 5 numbers each
A) 1,2,3,4,5 Sum = 15
B) 2,3,4,5,6 Sum = 20
C) 3,4,5,6,1 Sum = 19
D) 4,5,6,1,2 Sum = 18
E) 5,6,1,2,3 Sum = 17
F) 6,1,2,3,4 Sum = 16
Numbers formed from Set B,C,E,F satisfy our criteria (as sum is not a multiple of 3). Total codes formed from here = 4 * 5! = 480
Set A and D have a total which is multiple of 3.
Set A has 2 even numbers. So Codes formed (keeping even at units place) = 2*4! = 48
Set D has 3 even numbers. So Codes formed (keeping even at units place) = 3*4! = 72
Total Numbers = 480 + 48 + 72 = 600
A) 1,2,3,4,5 Sum = 15
B) 2,3,4,5,6 Sum = 20
C) 3,4,5,6,1 Sum = 19
D) 4,5,6,1,2 Sum = 18
E) 5,6,1,2,3 Sum = 17
F) 6,1,2,3,4 Sum = 16
Numbers formed from Set B,C,E,F satisfy our criteria (as sum is not a multiple of 3). Total codes formed from here = 4 * 5! = 480
Set A and D have a total which is multiple of 3.
Set A has 2 even numbers. So Codes formed (keeping even at units place) = 2*4! = 48
Set D has 3 even numbers. So Codes formed (keeping even at units place) = 3*4! = 72
Total Numbers = 480 + 48 + 72 = 600
