Question

27. Triplet x, y, and z are chosen from the set {1,2,3,.... 24,25} such that x < y < z. How many such triplets are
possible?
(a) 25c2
(b) 600
(c) 25c2, + 25c3
(d) 1200
​

Answer 1

Given :  Triplet x, y, and z are chosen from the set {1,2,3,.... 24,25} such that x < y < z.

To Find : How many such triplets are possible  

(a) 25c2

(b) 600

(c)  25c3

(d) 1200

Solution:

if x = 1

Then  y = 2   and  z can be 3 to 25   =  23 ways

      y = 3    and z can be     4 to 25  = 22 ways

     y = 24    and z can be   25       1  way

Sum = ( 1 + 2 + ______ 22 + 23)  =    23 * 24/2 =  

now if x = 2  then

          (1 + 2  + ______ + 21 + 22)  =  22 * 23 / 2

now if  x =  23     Then  only  case    =  1 * 2 / 2

so we can see this is

∑ (n)(n + 1)/2   where n is from 1 to 23

=  (1/2)  [ ∑ n²   + ∑ n  ]

= (1/2)  [  n(n+1)(2n+1)/6  +  n(n+1)/2 ]

=   (1/2)  (n(n+1)/2) [  (2n + 1)/3  +  1]

n = 23

=  (1/2)  (23(24)/2) [  (2(23) + 1)/3  +  1]

=  23 * 6 [ 47/3  + ]

= 23 * [ 94 + 6 ]

= 23 * 100

= 2300

=  ²⁵C₃

Simple method :  

we need to select   3 out of 25 which can be done

in ²⁵C₃   =  2300 ways

and then each  triplet can be arranged such that x  < y < z

Hence  ²⁵C₃   =  2300   such triplets are possible

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Answer 2

$\huge\red{\mathbb{Hello}}$

For Y = 1, there are 25 such doublets (1,1), (2,1), (3,1), . . . , (25,1).

For Y = 2, there are 24 such doublets (2,2), (3,2), . . . , (25,2).

For Y = 3, there are 23 such doublets (3,3), . . . , (25,3).

. . . . . . . . And so on . . . . .

Finally, for Y = 25, there is ONLY ONE such doublet (25,25).

So, the answer to your question is the value of the sum 25 + 24 + 23 + . . . + 1.

This value is very well known: it is the sum of the arithmetic progression

1 + 2 + 3 + . . . + 25 = %2825%2A26%29%2F2 = 325.