In how many ways can you select three integers x1, x2 &

Question

In how many ways can you select three integers x1, x2 & x3 from 1 to 12 such that x1<x2<x3. (Hint: Two ways to solve this. One lengthy...uhh and one very shorte... ahh)
 -Vicks

Praetorian · Answer

can u explain javs?

javropu · Answer

id do the following:
12*11*10/(3!)=220

Praetorian · Answer

Yeah..My answer cant be right.
But ...did you use 12C3 ?? 
Isnt 12C3 the total # of ways to draw 3 from 1-12 numbers without any restrictions?

Please explain

Praetorian

Vicky · Answer

right javs... 12C3 is the number of ways to select three numbers, and there is only one way to arrange these numbers such that x1 :!:

i hope u got it praet... -Vikcs

Praetorian · Answer

Clever way of phrasing the problem..same as the # of ways to pick 3 #'s from 1-12.. 

Thanks!

lakrana · Answer

if order does matter than why are we using a combination formula?


shouldn't we use the permutation formula?

Praetorian · Answer

If order matters ,we use permutations..
Permutations say the every position is unique..so 123 is different from 321...

But in our case...we dont need the positions to be unique..

hth
praetorian

Paul · Answer

In how many ways can you select three integers x1, x2 & x3 from 1 to 12 such that x1<x2<x3

venksune · Answer

can u give some answer choices...I think I am getting couple of different numbers here..

Paul · Answer

no answer choice in the original post, this one is just a very simple answer to a trickily worded question 8-)

You'll be blown away to know the answer... I already said too much.

Paul · Answer

Venksune, pick any 3 numbers 12C3 = 220 and there will ONLY ONE way of arranging them in increasing order

venksune · Answer

I just about wanting to correct to that...when I typeed my earlier message. Yes..I got that because the numbers are unique. right.

Anonymous · Answer

There are a total of 12C3=220 possibilities.

3 integers can be arranged in 3P3 possible ways i.e.3!/0!=6 ways of which only 1 possibility gives x1<x2<x3.

Hence the answer should be total no. of possibilities - total no. of unfavourable possibilities.

i.e. 220 - 5= 215.

Paul is my reasoning correct !!!

Epiphany · Answer

This perhaps might be a cumbersome way:

Let's keep 'a' fixed and move 'b'

a = 1, b = 2, then there are 12 - 2 = 10 possibilities for c such that a < b < c

a = 1, b = 3, then there are 12 - 3 = 9 possibilities for c such that a < b < c

a = 1, b = 4, then there are 12 - 4 = 8 possibilities for c

10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 5*11 = 55


Now when you move a to a = 2, and start b = 3 and move b up, the possibilities for c are


9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 - 10 = 45 (notice both series are the same except the first term, 10 is taken out)


Now when a = 3 b = 4, and you move b up then the total possibilities for c are 45 - 9 = 36, etc

55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 2 = 5*57 = 285

In how many ways can you select three integers x1, x2 &

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