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03 May 2020, 01:26
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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:
95%
(hard)
Question Stats:
30% (02:54) correct
70%
(02:43)
wrong
based on 105
sessions
History
Date
Time
Result
Not Attempted Yet
A ship has an area on its deck assigned to hold twelve private shipping containers in a row of six containers and up to two containers high. The regulations prohibit placement of a regular container on top of a refrigerated one. Due to construction limitations, there can be no other type of containers on top of a tank container, except itself. How many ways are there to stack six regular, three refrigerated, and three tank private shipping containers on the deck of the ship? (Each container belongs to a different owner, and thus is considered unique).
A) 6! x 3! x 3!
B) 6! x 4! x 214
C) 6! x 4! x 485
D) 6! x 6! x 74
E) 12!
A) 6! x 3! x 3!
B) 6! x 4! x 214
C) 6! x 4! x 485
D) 6! x 6! x 74
E) 12!
04 May 2020, 08:41
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lnm87
A calculation intensive question and not likely to be seen in the GMAT. However, the concept of combinations and Permutations do get tested..
If I get this question and I have to give a quick answer..
1) There are 12 places, so 12! ia the total ways. Since there are restrictions, we cannot have answer as 12!....Eliminate
2) No restriction on Regular ones, so if you place all 6 R down and the remaining 6 on top, the ways are 6!*6!..Eliminate A.
B=6!*4!*214=6!*6!*214/30~ 6!*6!*7, and similarly C=6!*6!*485/30~6!*6!*16....Pretty small as our answer in just one case is above 6!*6!..
We can make a wise guess as D..
Otherwise, the solution..
Various combinations..
Red-bottom row and blue-top row
1) All 6 R down, so other 3 Refrigated(A) and 3 T can placed on top ---- 6!*6!
2) Lower Row - 5R and 1A----(6C5*3C1*6!).....
Top-1R & 3T and 2A -----(5C1*5!).....Any of the 5 (3T+2A) on 1A in bottom row and the remaining 5 places in 5!
Total=(6C5*3C1*6!)*(5C1*5!)=6*3*6!*5*5!=6!*6!*15
3) Lower Row - 5R and 1T----(6C5*3C1*6!).....
Top-1R & 2T and 3A -----(2C1*5!)...Any of the 2T on 1T in bottom row and the remaining 5 places in 5!
Total=(6C5*3C1*6!)*(2C1*5!)=6*3*6!*2*5!=6!*6!*6
4) Lower Row - 4R and 1A and 1T----(6C4*3C1*3C1*6!).....
Top-2R & 2T and 2A -----(2C1*3C1*4!).....Any of the 2T on 1T in bottom row and remaining 3 (1T+2A) on 1A in bottom row and the remaining 4 places in 4!
Total=(6C4*3C1*3C1*6!)*(2C1*3C1*4!)=15*3*3*6!*2*3*4!=6!*6!*27
5) Lower Row - 4R and 2A ----(6C4*3C2*6!).....
Top-2R & 3T and 1A -----((3+1)C2*2!*4!).....Any of the 4 (3T+1A) on 1A in bottom row, and these can be placed in 2! ways and the remaining 4 places in 4!
Total=(6C4*3C2*6!)*(4C2*2!*4!)=15*3*6!*12*4!=6!*6!*18
6) Lower Row - 4R and 2T ----(6C4*3C2*6!).....
Top-2R & 1T and 2A ----NOT possible as T cannot have any other container on top and we have 2 T in bottom row while only 1 T in top row.
7) Lower Row - 3R and 1T and 2A ----(6C3*3C2*3C1*6!).....
Top-3R & 2T and 1A -----(2C1*2C2*2!*3!).....Any of the 2T on 1T in bottom row and remaining 2 (1T+1A) on 1A in bottom row, and these can be placed in 2! ways and the remaining 3 places in 3!
Total=(6C3*3C1*3C2*6!)*(2C1*2C2*2!*3!)=20*3*3*6!*2*2*3!=6!*6!*6
8) Lower Row - 3R and 3A ----(6C3*3C3*6!).....
Top-3R & 3T -----(3!*3!)..., 3T on 3A in bottom row and the remaining 3 places in 3!
Total=(6C3*3C3*6!)*(3!*3!)=20*6!*3!*3!=6!*6!
Total = \(6!*6!*(1+15+6+27+18+6+1)=6!*6!*74\)
04 May 2020, 23:20
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chetan2u
chetan2u
Thanks for an awesome explain. It is far far better than OE. I was banging my head but then i realized what you said - it can't be a GMAT type question.
However, i shared for the reason for it tests P&C concept.
I tried to find a short way to answer it by finding unit digit after eliminating A and E for reasons quite obvious but i faltered. Thank you once again - it is one of the best explanation i have seen in a while.
I will post OE in a while.
