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What could be the value of n, if the number of different codes that co
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06 Nov 2023, 00:16
06 Nov 2023, 00:16
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75% (01:59) wrong based on 4 sessions
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What could be the value of n, if the number of different codes that consist of 2 A’s and n B’s is less than 66?
Indicate all possible values.
(A) 2
(B) 3
(C) 5
(D) 9
(E) 10
(F) 12
Indicate all possible values.
(A) 2
(B) 3
(C) 5
(D) 9
(E) 10
(F) 12
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Official Answer
A,B,C,D
Re: What could be the value of n, if the number of different codes that co
[#permalink]
12 Nov 2023, 02:52
12 Nov 2023, 02:52
1
This is a good question to leverage the answer choices.
the total way that these letters can be arranged is: (2+n)!/2!*n! -> we have a total of 2+n As and Bs, with n Bs repeating and 2 As repeating.
Now we find the boundary:
With some logic we realise we should test D,E or F.
Try D first.
If n=9, then the equation becomes (11)!/2!*9! which equals 11*10/2=55.
So we are near the boundary.
To be sure, check E.
If n=10, (12)!/2*10!=12*11/2=66
Since the combinations are LESS than 66, n can be at most 9.
Therefore, A B C and D are valid options.
the total way that these letters can be arranged is: (2+n)!/2!*n! -> we have a total of 2+n As and Bs, with n Bs repeating and 2 As repeating.
Now we find the boundary:
With some logic we realise we should test D,E or F.
Try D first.
If n=9, then the equation becomes (11)!/2!*9! which equals 11*10/2=55.
So we are near the boundary.
To be sure, check E.
If n=10, (12)!/2*10!=12*11/2=66
Since the combinations are LESS than 66, n can be at most 9.
Therefore, A B C and D are valid options.
gmatclubot
Re: What could be the value of n, if the number of different codes that co [#permalink]
12 Nov 2023, 02:52
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