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Theorems on Combination

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Theorems on Combination - Lesson Summary

Theorem 1: The first theorem states the relation between the permutations and combinations of the n objects taken r at a time.

                      p r n   = C r n   × r!  
 
Suppose there are n objects and we have r places to be filled by taking r objects at a time.
 
The number of permutations is nP r. The process is of two events.
The first event is selecting r objects from the given n objects.
The second event is arranging these ‘r’ objects in ‘r’ places.
 
The first event of selecting r objects can be achieved in nC r ways.
 

 
The second event of arranging r items in r places can be done in r factorial ways.
 

Using the multiplication rule, the whole process is equal to nC r x r!

p r n   = C r n   × r!
 
Based upon the theorem, few results can be obtained.
 
Result 1:

C r n    = n! (n - r)!r!

C r n   × r! = P r n  

⇒ C r n × r! r! = P r n r!

⇒ C r n = P r n r!

⇒ C r n = n! (n - r)!r!

Result 2:

C 0 n = 1

⇒ C 0 n = n! (n - 0)!0!

         = n! (n - 0)!0!

         =  n! (n )! x 1   = 1    [ ∵ 0! = 1 ]


Result 3:

C n - r n = C r n

L.H.S :  n! (n - n + r)! × (n - r)!

          =  n! r! × (n - r)!   = C r n

Result 4:

C n n    = 1 

C n n    = C n - n n

C n n    = C 0 n   = 1

Result 5:

C a n    = C b n , then a = b or n = a + b

Given C a n    = C b n

Suppose a ≠ b

Let a < b

⇒ n! (n - a)!a! = n! (n - b)!b!

⇒ (n - a)!a! = (n - b)!b!

⇒ [(n - a) × (n - a - 1) × (n - a - 2)...(n - b + 1) × (n - b)! ] × a! = (n - b)! × [b × (b - 1) × (b - 2) × ...× 1]
 
The first term on the left hand side is expanded using the fact that a is less than b. The second term on the right hand side is also expanded in a similar way.
 
⇒ (n - a)(n - a - 1)(n - a - 2)...(n - b + 1) = b(b - 1)(b - 2)...(a + 1)
 
Since either side is the product of ‘b - a’ consecutive positive integers, we have ‘n minus a’ is equal to b.
 
⇒ n - a = b

⇒ n = a + b
 
Theorem 2
 
C r n + C r - 1 n = C r n + 1

L.H.S = n! (n - r)!r! + n! (n - r + 1)!(r - 1)!

         = [(n - r)! × (r × (r - 1)!)] + n! [(n - r + 1) × (n - r)!] × (r - 1)!

         = n! (n - r)!(r - 1)! [ 1 r + 1 (n - r + 1) ]

         = n! (n - r)!(r - 1)! [n - r + 1 + r r x (n - r + 1) ]

         = n! (n - r)!(r - 1)! [n + 1 r x (n - r + 1) ]

         = (n + 1) × n! [r × (r - 1)!] × [(n - r + 1) × (n - r)!]
 
         = (n + 1) ! r ! × [(n - r + 1)!]

        = (n + 1) ! r ! × [(n + 1 - r)!]
 
        = C r n + 1
 
Hence the theorem is proved.

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