Combinations - Formula for ⁿC_r

The number of combinations of n different things taken r at a time is
                   ⁿC_r = n!/[ r! (n -r)!]

Corollaries:

1. ⁿC_r = [n (n -1)(n -2).....upto r factors]/r!
2. ⁿC_n = 1
3. ⁿC₀ = 1
4. ⁿC_r = ⁿC_{n -r}

Illustrative Examples

Example

Prove that ⁿC_r +ⁿC_{r -1} =^{n +1}C_r.

Solution

ⁿC_r +ⁿC_{r -1} = n!/[ (n -r)! r!] + n!/[(n -(r -1))! (r -1)!]

=

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

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

= (n +1)!/[r! (n +1 -r)!] = ^{n +1}C_r

Example

Evaluate

1. C(12, 5)
2. ¹⁰C₈
3. ¹⁰C₇ +¹⁰C₆
4. ¹⁵C₈ +¹⁵C₉ -¹⁵C₆ -¹⁵C₇

Solution

1. C(12, 5) = 12!/(5! 7!) = 792
2. ¹⁰C₈ = ¹⁰C₂ = (10.9)/(1.2) = 45
3. ¹⁰C₇ +¹⁰C₆ = ¹¹C₇       (using ⁿC_r +ⁿC_{r -1} =^{n +1}C_r)
         = ¹¹C₄ = (11. 10. 9. 8)/(1. 2. 3. 4) = 330      (using ⁿC_r = ⁿC_{n -r})
4. ¹⁵C₈ +¹⁵C₉ -¹⁵C₆ -¹⁵C₇  = (¹⁵C₈ +¹⁵C₉) -(¹⁵C₆ +¹⁵C₇ )
           =¹⁶C₉ -¹⁶C₇ = 0    (because¹⁶C₉ =¹⁶C₇)

Exercise

1. Evaluate the following:
   (i) C(15, 11)
   (ii) ¹⁰⁰C₉₈
   (iii) ⁵²C₅₂
   (iv) C(11, 7) -C(10, 6)
   (v) ⁷C₄ +⁷C₅ +⁸C₆
   (vi) ⁵C_r.
2. Prove that (i) ⁿC_r +2.ⁿC_{r -1} +ⁿC_{r -2} = ^{n +2}C_r
   (ii)
   (iii) n.^{n -1}C_r-1 = (n -r +1).ⁿC_r-1
3. (i) If ¹⁶C_r = ¹⁶C_{r +2} find ^rC₄
   (ii) If ⁿC₂ = ⁿC₃ find ⁿC₅.
   (iii) If ²ⁿC₃ : ⁿC₃ = 11 : 1 find n.
   (iv) If ⁿP_r = 720 and ⁿC_r = 120 find r.
4. Find the value of ⁴⁷C₄ +^{(52 -r)}C₃
5. If ⁿC_{r -1} = 36, ⁿC_r = 84 and ⁿC_{r +1} = 126, find the values of n and r.

Answers