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19 Nov 2019, 03:23
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
54% (01:49) correct
46%
(02:10)
wrong
based on 248
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What is the sum of all possible 7-digit numbers that can be constructed using the digits 1, 2, 3, 4, 5, 6 and 7, if each digit can be used only once in each number?
A) 7! * 8888888
B) 3.5! * 8888888
C) 3.5!/2 * 88888888
D) 7!/2 * 8888888
E) 7 * 8888888
A) 7! * 8888888
B) 3.5! * 8888888
C) 3.5!/2 * 88888888
D) 7!/2 * 8888888
E) 7 * 8888888
firas92
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20 Nov 2019, 23:22
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This is a fairly common question type in the GMAT and there is even a formula to solve this
1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: (n-1)!*(Sum of the digits)*(1111....n times)
2. Sum of all the numbers which can be formed by using the n digits with repetition being allowed is: \(n^{n-1}\)*(Sum of the digits)*(1111....n times)
In this question, we are not allowed repetition and so the answer is
\((7-1)!*(1+2+..+7)*11111111\)
That is, \(6!*28*1111111 = 6!*7*4*1111111 = 7!*4*1111111 = \frac{7!}{2}*2*4*1111111 = \frac{7!}{2}*8888888\)
Answer is (D)
1. Sum of all the numbers which can be formed by using the \(n\) digits without repetition is: (n-1)!*(Sum of the digits)*(1111....n times)
2. Sum of all the numbers which can be formed by using the n digits with repetition being allowed is: \(n^{n-1}\)*(Sum of the digits)*(1111....n times)
In this question, we are not allowed repetition and so the answer is
\((7-1)!*(1+2+..+7)*11111111\)
That is, \(6!*28*1111111 = 6!*7*4*1111111 = 7!*4*1111111 = \frac{7!}{2}*2*4*1111111 = \frac{7!}{2}*8888888\)
Answer is (D)
19 Nov 2019, 08:12
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Yes, rocksailor9,
The question is likely feasible in GMAT exam, if we correlate the sequence property here.
This type of question does not follow directly the Arithmetic properties but indirectly we can apply, i.e. The difference of each side of the AP sequence have the same difference from the mean. e.g. 123, 132, 213, 231, 312, 321 has the mean as 222. Now if you find the difference between each term with respect to mean (123, 132, 213, 222, 231, 312, 321), the first and last must have same difference. Second last of each side must have same difference. The same pattern follows here also. WE CAN APPLY THE SAME PROPERTY TO N NUMBER OF DIGITS. So, we can use the formula to find sum of A.P. sequence.
Sum = n/2 (a+l)
Now, n = Total possible no. of numbers = 7p7 = n! = 7!
a= first number, i.e. smallest one = 1234567
l= last number, i.e. largest one = 7654321,
= (7!/2) (1234567+7654321)
= 7!/2 * 88888888
Ans. D.
Hope this helps.
Bunuel, kindly let me know any other better and quick approach.
The question is likely feasible in GMAT exam, if we correlate the sequence property here.
This type of question does not follow directly the Arithmetic properties but indirectly we can apply, i.e. The difference of each side of the AP sequence have the same difference from the mean. e.g. 123, 132, 213, 231, 312, 321 has the mean as 222. Now if you find the difference between each term with respect to mean (123, 132, 213, 222, 231, 312, 321), the first and last must have same difference. Second last of each side must have same difference. The same pattern follows here also. WE CAN APPLY THE SAME PROPERTY TO N NUMBER OF DIGITS. So, we can use the formula to find sum of A.P. sequence.
Sum = n/2 (a+l)
Now, n = Total possible no. of numbers = 7p7 = n! = 7!
a= first number, i.e. smallest one = 1234567
l= last number, i.e. largest one = 7654321,
= (7!/2) (1234567+7654321)
= 7!/2 * 88888888
Ans. D.
Hope this helps.
Bunuel, kindly let me know any other better and quick approach.
