+0

# PL5 Math Question 8

+1
55
6
+134

A certain cryptocode must contain one letter from the set {X, K, M, Z} and three distinct letters from the set {W, X, Y, Z}. The four letters can be arranged in any order, and since X and Z are in both sets, these letters may each appear twice in an arrangement. How many cryptocodes are possible?

dgfgrafgdfge111  Nov 22, 2018
#1
+456
+1

The set {X, K, M, Z} has 4 possible letters, so we can pick any 4.

The set {W, X, Y, Z} has 4 possible letters, but since we pick 3 out of 4, there are $$4\times 3\times 2=24$$.

$$24\times 4 = 96$$ crypto codes

You are very welcome!

:P

CoolStuffYT  Nov 22, 2018
#2
+3213
+1

I'm afraid this isn't correct.

Ignoring duplicates for a moment there are

$$n = \dbinom{4}{1}\dbinom{4}{3} = 16$$ different codes that are generated as described

There is however 1 code that is duplicated

(w,x,y,z) can be obtained by either (x)+(w,y,z) or (z) + (w,x,y)

So we end up with a total of 15 distinct codes.

Rom  Nov 23, 2018
#3
+2

HERE WE ARE ..

If X is from the first set and from the other set W, X, Y are chosen, then possible cryptocodes are:

XWXY, XWYX, XXWY, XXYW, XYXW, XYWX, WXXY, WXYX, WYXX, YWXX, YXWX, YXXW - altogether 12 possibilities

If X is from the first set and from the other set we don't chose X at all, then we have 24 possibilities, for example:

if we chose X from the first set and W, Y, Z from the other set, we can have the following cryptocodes:

XWYZ, XWZY, XYWZ, XYZW, XZWY, XZYW, WXYZ, WXZY, WYXZ, WYZX, WZXY, WZYX,

YWXZ, YWZX, YXWZ, YXZW, YZXW, YZWX, ZWXY, ZWYX, ZYWX, ZYXW, ZXWY, ZXYW

We can repeat until all letters have been chosen.

Every time when the four chosen letters are different, we have 24 possibilities, and if there are some letters twice, then we have 12 possibilites.

Result: there are altogether 312 possibilties!

Isn't it?

Guest Nov 23, 2018
edited by Guest  Nov 23, 2018
#4
+3213
+2

The problem states that the letters can be arranged in any order.

I take that to mean that (w,x,y,z) and (x,y,z,w) and all it's permutations are the same code.

Rom  Nov 23, 2018
#5
+2

Then I have to ask, what it means - cryptocode?

I suggested that ABCD is not equal to any of permutations (ABDC, BACD and so on).

Guest Nov 23, 2018
#6
+94183
+2

The question is open to interpretation.

Personally I think that when it says the letters can be arranged in any order it means that they can be arranged in any order and each one of those orders counts as a different combination.

So lets see if I can get an answer given my interpretation.

A certain cryptocode must contain one letter from the set {X, K, M, Z} and three distinct letters from the set {W, X, Y, Z}. The four letters can be arranged in any order, and since X and Z are in both sets, these letters may each appear twice in an arrangement. How many cryptocodes are possible?

 no of combinations no of permutations each Total permutations X X plus 2 of 3 3 4!/2!=12 3*12=36 X No X 1 4!=24 1*24=24 Y includes Y 3 12 3*12=36 Y No Y 1 24 1*24=24 Not X or y Any 3 2*4=8 4!=24 8*24=192 Total 312

If my interpretation is correct I get 312 possible crypocodes.

Melody  Nov 27, 2018