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29 Apr 2025, 00:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
59% (01:47) correct
41%
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wrong
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Let N = {-6, 0, 5, 9, 14, 12, 8, 22}. Then the average of the average of all the possible triplets formed from N is
A. 7
B. 8
C. 9
D. 10
E. 11
A. 7
B. 8
C. 9
D. 10
E. 11
29 Apr 2025, 06:03
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Is the question wrongly written? What does average of average of all possible triplets mean? Bunuel
I think it's just average of all possible triplets.
Take an easier example :
{1,2,3,4}
4C3 = 4
1,2,3
1,2,4
1,3,4
2,3,4
If you sum them up it's 3(1) + 3(2) + 3(3) + 3(4) = 3(1+2+3+4)
Similarly, sum of all triplets of the given set is 7(sum of all elements is the set) = 7 * 64
Number of triplets = 8C3 = 8!/3!5! = 8 * 7 = 56
Average of all triplets = Sum of all triplets/number of triplets = 7 * 64/56 = 8
I think it's just average of all possible triplets.
Take an easier example :
{1,2,3,4}
4C3 = 4
1,2,3
1,2,4
1,3,4
2,3,4
If you sum them up it's 3(1) + 3(2) + 3(3) + 3(4) = 3(1+2+3+4)
Similarly, sum of all triplets of the given set is 7(sum of all elements is the set) = 7 * 64
Number of triplets = 8C3 = 8!/3!5! = 8 * 7 = 56
Average of all triplets = Sum of all triplets/number of triplets = 7 * 64/56 = 8
29 Apr 2025, 22:19
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Bunuel
This is an amazing question to clarify certain concepts.
N = {-6, 0, 5, 9, 14, 12, 8, 22}. The total number of elements is 8.
How many triplets are formed from these 8 elements ?
8C3 =( 8*7*6)/(3*2*1) = 56 groups of three element set is formed.
Example to understand concept :
Lets first consider an example of a five element set {1 , 2, 3, 4, 5 }
Number of triplets formed = 5c3 = 5c2 = 10
The possible 10 triplets are :
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
The number of occurrence of each digit is 6 times. To form a triplet , after choosing a digit say 1, we are left with 4 digits to choose 2 letters from it, which is 4c2 = 6.
6*(1)+6*(2)+6*(3)+6*(4)+6*(5) = 6+12+18 + 24+ 30 = 90
since each digit appears 6 times, 6*5 = 30.
90/30 = 3.
For the five element set {1 , 2, 3, 4, 5 }. Sum = 15, number of elements = 5, average = 3.
to verify it with another extra element, if the pattern works .
For a 6 element set , sum = 21, number of element =6 , average = 3.5
6*(1)+6*(2)+6*(3)+6*(4)+6*(5)+ 6*(6) = 6+12+18 + 24+ 30+36 = 126
since each digit appears 6 times, 6*6 = 36.
126/36 = 3.5
Now to the original problem,
N = {-6, 0, 5, 9, 14, 12, 8, 22}, re arranging it
Sum = -6+0+5+9+14+12+8+22 = 64
number of elements = 8
average = 64/8 = 8
Alternate approach:
triplets from 8 elements = 56
each Digit appears 21 times. After selecting a digit, we have 7c2 ways to select the remaining = 21 ways. So, 1 occurs 21 times.
21*(-6)+ 21*(0)+21*(5)+21*(9)+21*(14)+21*(12)+21*(8)+21*(22)
= -126 + 0+ 105+ 189+ 294+252+168+462
= 1344
each digit appears 21 times. So for 8 digit = 21*8 = 168
Average = 1344/168 = 8
Option B
