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09 Dec 2021, 07:00
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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:
95%
(hard)
Question Stats:
40% (03:07) correct
60%
(02:33)
wrong
based on 47
sessions
History
Date
Time
Result
Not Attempted Yet
How many distinct eight-character codes-with all characters distinct, containing only two numerals can be formed from A, B, C, D, E, F, 1, 2, 3, 4, such that all of them start with B and no two letters of B, D, F are together?
A. 2400
B. 14400
C. 36000
D. 43200
E. 44300
Are You Up For the Challenge: 700 Level Questions
A. 2400
B. 14400
C. 36000
D. 43200
E. 44300
Are You Up For the Challenge: 700 Level Questions
09 Dec 2021, 07:23
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We have 4C2 = 6 ways of choosing the two numerals, and we need to use all six letters. So if we imagine we've picked our two numerals already, and count our possibilities, we'll then need to multiply by 6 to get the answer, to account for the different choices we could make of the numerals. If we've chosen our eight characters, we have 1 choice for our first character (it must be 'B'), and then if there were no restrictions at all, we could make 7! codes, and the answer would be 6*7!. That's just slightly bigger than 30,000, so answers C, D and E are all impossible, and the restriction will reduce the number of codes, but not by more than 90%, so only answer B is remotely plausible.
If we want to solve exactly, the first character is 'B', and the second cannot be 'D' or 'F', so we have 5 choices for the second character. We then have six slots left to fill. With no restrictions, there would be 6C2 = 15 choices we could make for the two slots we'll put the 'D' and 'F' in. Since they can't be together, we must subtract the 5 pairs of adjacent slots, so we have 10 options for the two slots where we can put 'D' and 'F', and those two letters can be in either order, for 2*10 possible places for those two letters. We then can freely put our last four symbols in the last four slots, so we have 4! possibilities for the last four characters. So the answer is, multiplying all of our choices from every step, 6*5*20*4! = 6*100*24 = 14,400. This problem is quite a bit more cumbersome than any similar official question I've seen, though.
If we want to solve exactly, the first character is 'B', and the second cannot be 'D' or 'F', so we have 5 choices for the second character. We then have six slots left to fill. With no restrictions, there would be 6C2 = 15 choices we could make for the two slots we'll put the 'D' and 'F' in. Since they can't be together, we must subtract the 5 pairs of adjacent slots, so we have 10 options for the two slots where we can put 'D' and 'F', and those two letters can be in either order, for 2*10 possible places for those two letters. We then can freely put our last four symbols in the last four slots, so we have 4! possibilities for the last four characters. So the answer is, multiplying all of our choices from every step, 6*5*20*4! = 6*100*24 = 14,400. This problem is quite a bit more cumbersome than any similar official question I've seen, though.
02 Feb 2022, 02:33
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Bunuel
The code will be Bpqrstuv.
Numerals: We have to choose 2 out of 4. => 4C2 or 6 ways.
Now q cannot be D or F, so q can be filled in 5 ways.
Remaining 6 places can be filled in 6! ways.
Thus, without any further restrictions, total ways are 6*5*6! or 21600.
BUT, the above contains places where D and F are together.
Take D and F as 1, and they can be arranged in 2 ways.
As above q can be filled in 5 ways, and remaining (6-1) or 5 ways can be filled in 5! ways.
Total ways: 6*5*5!*2=7200 ways.
Required answer = 21600-7200 = 14400
B
