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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote how many different arrangements can the squares have... tagged by: BTGmoderatorLU This topic has 2 member replies Top Member how many different arrangements can the squares have... Each of the squares above is to have exactly one letter and nothing else placed inside it. If 3 of the letters are to be the letter X, 2 of the letters are to be the letter Y, and 1 of the letters is to be the letter Z, in how many different arrangements can the squares have letters placed in them? A. 30 B. 60 C. 108 D. 120 E. 720 The OA is B. I'm really confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance. Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 320 messages Upvotes: 27 LUANDATO wrote: Each of the squares above is to have exactly one letter and nothing else placed inside it. If 3 of the letters are to be the letter X, 2 of the letters are to be the letter Y, and 1 of the letters is to be the letter Z, in how many different arrangements can the squares have letters placed in them? A. 30 B. 60 C. 108 D. 120 E. 720 The OA is B. I'm really confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance. Well, pretend for the moment that each of the letters is different, even though we know they are not all distinct. So with 6 "different" letters in mind, place a letter in the first box. There are 6 ways to do this. Place a letter in the second box from among the 5 remaining letters, so 5 ways to do this. And so on. So there are 6x5x4x3x2x1 arrangements. Now let's look at one possible pattern: XXXYYZ. Looking at the 3 Xs remember that the above calculation treats each X as different from another X, which we know isn't correct. If you had 3 distinct things in the 3 spots occupied by the Xs there would be 3x2x1 ways of arranging them, just as with our first calculation. Likewise, there are 2 Ys being treated as different which corresponds to 2x1 ways of arranging. Since this is multiplication you need to reduce the original calculation using division as follows to adjust for the above: (6x5x4x3x2x1)/((3x2x1)x(2x1)) = 5x4x3 = 60 Senior | Next Rank: 100 Posts Joined 15 Jan 2018 Posted: 83 messages X can go in any 3 of 6 boxes, so there are 20 choices for X as follows: XXX000 XX0X00 XX00X0 XX000X X0XX00 X0X0X0 X0X00X X00XX0 X00X0X X000XX 0XXX00 0XX0X0 0XX00X 0X0XX0 0X0X0X 0X00XX 00XXX0 00XX0X 00X0XX 000XXX Of the 3 boxes remaining, for each of the above, Y can go in any 2 of the 3 boxes. So there are 3 choices for Y: YY0 Y0Y 0YY As for Z, there is only 1 space remaining in each scenario above, so it has no choice. Therefore the number of different arrangements is 20 x 3 = 60. • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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