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  • nCr , nCn =1, nC0 = 1, nCr = nCn–r, nCx = nCy, then x + y = n or x = y, n+1Cr = nCr-1 + nCr
  • When all things are different
  • When all things are not different.
  • Mixed problems on permutation and combinations.


Theorem: `"^n P_r`= `"^n C_r` r!, 0 < r ≤ n.
Proof:  Corresponding to each combination of `"^nC_r`, we  have r ! permutations, because r objects in every combination can be rearranged in r ! ways. 
Hence, the total number of permutations of n different things taken r at a time is `"^nCr` × r!. On the other hand, it is P n r . Thus
`"^n P_r` =`"^n C_r` * r!, 0 < r ≤ n.


1)  From above n!/(n-r)!= `"^n C_r` * r!, i.e., `"^n C_r`= n!/[r!(n-r)!]
In particular, if r= n, `"^n C_n`= n!/(n!0!)= 1
2) We define `"^nC_0` = 1, i.e., the number of combinations of n different things taken nothing at all is considered to be 1. Counting combinations is merely counting the number of ways in which some or all objects at a time are selected. Selecting nothing at all is the same as leaving behind all the objects and we know that there is only one way of doing so. This way we define `"^nC_0` = 1.
3) As `(n!)/[0!(n-0)!]`= 1= `"^nC_0`, the formula `"^n C_r`= `(n!)/[r!(n-r)!]` is applicable for r=0 also. Hence
`"^n C_r`= `(n!)/[r!(n-r)!], 0 < r ≤ n`.
4) `"^n C_n-r`= `(n!)/ [(n-r)! (n-(n-r))!]= (n!)/[(n-r)!r!]= ``"^n C_r`, 
i.e., selecting r objects out of n objects is same as rejecting (n – r) objects. 
5) `"^nC_a` = `"^nC_b` ⇒  a = b  or a = n – b, i.e., n = a + b 


Theorem: `"^nC_r` + `"^nC_r-1`= `"^(n+1)C_r` 
Proof:  We have `"^nC_r` + `"^nC_r-1= (n!)/[r!(n-r)!] + (n!)/[(r-1)!(n-r+1)!]`

= `(n!)/ [r*(r-1)!(n-r)!] + (n!)/[(r-1)!(n-r+1)(n-r)!]`

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

= `(n!)/[(r-1)!(n-r)!] * (n-r+1+r)/[r(n-r+1)]`

= `(n+1)!/[r!(n+1-r)!`

= `"^(n+1) C_r`

If you would like to contribute notes or other learning material, please submit them using the button below. | Theorem: `"^nC_r` + `"^nC_r-1`= `"^(n+1)C_r`

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Theorem: `"^nC_r` + `"^nC_r-1`= `"^(n+1)C_r` [00:06:31]

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