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## 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.

### Top Member

Master | Next Rank: 500 Posts
Joined
15 Oct 2009
Posted:
308 messages
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.

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