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

1. 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.

1. 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-(2a-1)
similarly for B & C
So , [ 76 - (2a-1) ]+ [ 76- (2b-1)]+[76-(2c-1)]= 153
2(a+b+c)+3=76(3) - 153
a+b+c= 36
total (36+3-1)C(3-1) = 38C2 ways = 703 ways
when a=b
2a+c= 36
a goes from 0 - 18 but this includes case where a = b = c
so 19-1 = 18 cases
similarly 18 cases for b = c & c = a each
and 1 case of equilateral triangle

Scalene - > (703-(38*3) - 1) /6 = 648/2 = 108
isosceles - > 18
equilateral - > 1
total 108+18+1 = 127

Or 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

• Find the number of triangle having odd integral sides of perimeter equal to 153

what would be the direct formula for this @gaurav_sharma

• @cat789

Number of Triangles with Integer sides for a given perimeter.

• If the perimeter p is even then, total triangles is [p^2]/48.
• If the perimeter p is odd then, total triangles is [(p+3)^2]/48
• If it asks for number of scalene triangle with a given perimeter P, then subtract 6 and apply the same formula . For even [(p-6)^2]/48 and for odd [(p-3)^2]/48.

Where [x] represents neatest integer function. For example [6.7] is 7 not 6 because its nearest integer.

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