Hi there!
Phase 1: Number of (x,y,z) triplets such that x<y<z :
10*9*8 over 3! = 6, therefore 120 cases.
Phase 2: Number of (x,y,z) triples such that x = y < z:
If z = 10, we have from x=y=1 to x=y= 9, therefore 9 cases.
If z = 9, we have from x=y=1 to x=y=8, therefore 8 cases.
...
If z = 2, we have x=y=1, therefore 1 case.
Total of Phase 2 cases: 1+2+...+9 = 9*((1+9)/2) = 45 cases.
Phase 1 and Phase 2 are mutually exclusive, therefore 120+45 = 165 cases.
Regards,
Fabio.
Number of triplets
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fskilnik@GMATH
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Rahul@gurome
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Number of triplets (x ≤ y < z) = Number of triplets (x < y < z) + Number of triplets (x = y < z)oddball wrote:Tiplet (x, y, z) is chosen from the set {1, 2, 3, ... , 10}such that x ≤ y < z. How many such triplets are possible?
(A) 120
(B) 165
(C) 330
(D) 810
(E) 1200
Number of triplets (x < y < z) = Number of ways to select 3 integers out of 10 = 10C3 = 120
Number of triplets (x = y < z) = Number of doublets (x < z) = Number of ways to select 2 integers out of 10 = 10C2 = 45
Thus, number of triplets (x ≤ y < z) = (120 + 45) = 165
The correct answer is B.
Rahul Lakhani
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fskilnik@GMATH
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I´ve tried to avoid this approach because I believe it is a bit "harder-than-necessary", in the sense that we must "justify" (at least to ourselves) why this way of counting is correct... well, let´s do it!Rahul@gurome wrote: Number of triplets (x < y < z) = Number of ways to select 3 integers out of 10 = 10C3 = 120
Number of triplets (x = y < z) = Number of doublets (x < z) = Number of ways to select 2 integers out of 10 = 10C2 = 45
When you select 3 integers out of 10 BY COMBINATION, you mean you are counting (for instance) the 3! = 6 selections (1,2,3) , (2,1,3), ... , (3,2,1) as just one or, in other words, we are selecting one "representative" from these 6 possibilities.
The "trick" (and beauty) is the fact that we may ASSUME that the choice was exactly the one we are looking for, that is, the only one for which we have x<y<z ....
The same argument is necessary to justify Rahul´s selection BY COMBINATION of 2 integers out of 10... but now we have also the task to "justify" (at least to ourselves) why we don´t need to "bother" with the third choice... the reason is because Rahul is "admitting" that he will choose x and z, in the DESIRED order x < z (as explained in the previous paragraph), and that y will be taken as the same value of x !! In other words, y has "no choice" but to be the same as x, therefore we don´t need to "bother" counting y as far as we are counting x successfully.
Well, I hope I was understood. That´s certainly a beautiful approach to the problem but I guess it´s a bit more risky (to be found) by the average student.
Regards,
Fabio.
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