Permutation & Combination concepts by Gaurav Sharma  Part (5/5)

Find the total number of triangles with integral sides and perimeter 122
Let A, B and C be the sides of a triangle.
A + B + C = 122
Also, A + B > C, B + C > A, A + C > B (sum of two sides > third side)
So maximum value of any side will be (122/2) – 1 = 60
S0, (60 – a) + (60 – b) + (60 – c) = 122 [0 ≤ a, b, c ≥ 59]
a + b + c = 58
So, total cases will be (58 + 3 – 1) C (3 – 1) = 60C2 = 30 x 59
This will include cases where ( a = 19, b = 18, c = 21) , ( a = 18, b = 19, c = 21) … but all of the cases are forming the same triangles. So we need to subtract the repetitions.
Now, a + b + c = 58
 Where any two of a, b and c are the same.
2a + c = 58, here a can take values from 0 to 29 so 30 cases.
Similarly for b = c and c = a hence total 30 x 3 = 90 cases.
 When a = b = c, not possible as a + b + c = 58
So, cases where all are different will be (59 x 30) – 90 = 1680
For distinct cases we will divide this by 3! = 6
We will get 1680/6 = 280
Cases where two are the same = 90
For distinct cases we will divide this by 3, 90/3 = 30
Total cases = 280 + 30 = 310
Short cut:
Total number of triangles with integral sides and perimeter n
 When n is odd : [ (n + 3)^2/48]
 Where n is even : [n^2/48]
Where [.] is the nearest integer (Don’t confuse with greatest integer function)
Applying for the above problem, [122^2]/48 = [310.08] = 310
To find the number of scalene triangles with integral sides and perimeter n
Put ( n – 6) instead of n in the above formula (for all triangles)
 When n is Odd : [(n – 3)^2/48]
 When n is even : [(n – 6) ^2/48]
Where [.] is the nearest integer
Example: Perimeter of a triangle is 150. Find the number of scalene triangles possible with integral sides
Using short cut – n = 150 (even)
Number of scalene triangles = [(150 – 6)^2/48] = 432
Detailed approach:
A + B + C = 150. Max value of any side can be 74
So, (74 – a) + ( 74 – b) + (74 – c) = 150
a + b + c = 72
Total cases ( 72 + 3 – 1) C ( 3 – 1) = 74C2 = 37 x 73
When two of them are same:
2a + c = 72, where a goes from 0 to 36 but if a = 24 then b = c = a
So 36 cases
Similarly 36 cases for b = c and c = a each
Also where a = b = c = 24  > 1 case
(37 x 73) – ( 36 x 3) – 1 = 2592
Now divide this by 3! = 6 to find distinct cases
Number of scalene triangles = 2592/6 = 432
How many scalene triangles are possible if all the sides are integers and the perimeter of a triangle is 24 units?
OA: 7
Make sure that it is not greatest integer function here. We have to take nearest value
The perimeter of a triangle is 150. Find the number of isosceles triangles possible with integral sides.
A + B + C = 150
Max value of any side can be 74
So, (74 – a) + ( 74 – b) + (74 – c) = 150
a + b + c = 72
Total cases ( 72 + 3 – 1) C ( 3 – 1) = 74C2 = 37 x 73
When two of them are same
2a + c = 72, where a goes from 0 to 36 but if a = 24 then b = c = a
So 36 cases
Similarly 36 cases each for b = c and c = a each
Total 36 x 3 = 108 triangles.
But we have to exclude the repetitions
So 108/3 = 36 triangles.
Short cut:
To find the number of isosceles triangles with integral sides and perimeter n
No. of total triangles – no. of scalene triangles – no. of equilateral triangles
Here perimeter = 150
Total triangles = [n^2/48] = [468.75] = 469
Scalene triangles = [(n – 6)^2/48] = [432] = 432
Isosceles triangles = 469 – 432 – 1 = 36
The number of integral even sided triangle with perimeter 180
A + B + C = 180
Max value of any side can be 89
But sides are even so max value will be 88
Also, A, B and C are even so put A = 88 – 2a, B = 88 – 2b, C = 88 – 2c
(88 – 2a) + (88 – 2b) + (88 – 2c) = 180
2(a + b + c) = 84
a + b + c = 42
Total cases : (42 + 3 – 1)C(3 – 1) = 44C2 = 946
When two of them are same:
2a + c = 42, where a goes from 0 to 21 but if a = 14 then a = b = c. so, 21 cases
Similarly 21 cases for b = c and c = a each
Also, where a = b = c = 24 – 1 case
946 – ( 21 x 3) – 1 = 882
Divide by 3! To get distinct cases
Number of scalene triangles = 2592/6 = 147
Number of isosceles triangles = 21 x 3 / 3 = 21
Number of equilateral triangles = 1
Total 147 + 21 + 1 = 169
Short cut:
Number of triangles with all sides (integral) EVEN
Total number of triangles with integral sides and perimeter n (n is even) = n^2/48
[When all sides are even perimeter will also be even]
But, here we have to find number of triangles with all sides even so put n/2 instead of n in the above formula
Required number of triangles – n^2/(4 x 48 )
Here, perimeter = 180 and all sides should be even
Number of triangles = [n^2/( 4 x 48 )] = [180^2/ ( 4 x 48 )] = [168.75] = 169
Find the number of triangle having odd integral sides of perimeter equal to 153
A+B+C=153
max value can be 76
put A = 76(2a1)
similarly for B & C
So , [ 76  (2a1) ]+ [ 76 (2b1)]+[76(2c1)]= 153
2(a+b+c)+3=76(3)  153
a+b+c= 36
total (36+31)C(31) = 38C2 ways = 703 ways
when a=b
2a+c= 36
a goes from 0  18 but this includes case where a = b = c
so 191 = 18 cases
similarly 18 cases for b = c & c = a each
and 1 case of equilateral triangleScalene  > (703(38*3)  1) /6 = 648/2 = 108
isosceles  > 18
equilateral  > 1
total 108+18+1 = 127Or Apply formula (n + 3)^2/48 = [126.75] = 127
The possible shortest routes in which we can travel from A to B (grid of m x n) are ( m + n )! / m!n!
The possible shortest routest in which we can travel from A to B where there is a shortcut are:
For example in grid of ( 6 x 4 )
From A to C in 4!/2!2! = 6 ways
From C to D = 1 way
From D to B = 4!/1!3! = 4 ways
Total ways = 6 x 1 x 4 = 24 ways