Combinatorial Learning Formula

Combination are selections ie. it involves only the selection of the required number of things out of the total number of things. Thus in combination order does not matter. For example, consider a set of three elements a,b,c and combination
made out of the set with


i) One at a time: a, b, c
ii) Two at a time: a,b, b,c, c,a
iii) Three at a time: a,b,c

The number of combinations of n things taken r, (r < n) is denoted by nCr or ([n],[r])

Learning the evaluation of combinations and the combinatorial formula

Learning to derive the formula for nCr:( Combinatorial formula)
No of combinations from ‘n’ things taken ‘r’ at a time = nCr
No of permutations from ‘n’ things taken ‘r’ at a time = nPr
Number of ways ‘r’ things that can be arranged among themselves = r!
Each combination having r things gives rise to r! permutations

Therefore  nPr = (nCr) r!


=> ((n - r )!) / (n!) = ( nCr) r!


nCr = r!(n - r )!

Learning observations related to combinatorial formula:

(i)   nCo = (n!)/(0! (n - 0 )!) = (n!)/(n!) = 1


(ii) nCn = (n!)/(n! (n -n)!) = (n!)/(n!0!) = 1


(iii) nCr = nCn-r


(iv) If nCx = nCy then x = y or x+y = n


(v) nCr =(^nPr)/(r!)

Learning examples on combinatorial formula

Learning to solve examples using the combinatorial formula :

Ex 1:Evaluate 8C3

Solution:

8C3  =  (8!)/(3! (8 - 3)!) = (8!)/(3! . 5!) = (8 . 7 . 6 . 5!)/((3 . 2 . 1) . 5!)  =  (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

OR another method

8C3 = (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

Ex 2:

Evaluate: 9C7

Solution:

9C7 = 9C2 = (9 . 8)/(2 . 1) = 9 . 4 = 36

Ex 3:    A committee of seven students is formed selecting from 6 boys and 5 girls such that majority are from boys. How many different committees can be formed?

Sol:

Number of students in the committee = 7
Number of boys = 6
Number of girls = 5
The selection can be done as follows


Boy (6)    Girl (5)
6              1
5              2
4              3


ie. (6B and 1G) or (5B and 2G) or (4B and 3G)

The possible ways are ([6],[6]) ([5],[1]) or ([6],[5]) ([5],[2 ]) or ([6],[4]) ([5],[3])

The total number of different committees formed

=  6C6  x  5c1 + 6C5  x  5C2  +  6C4  x  5C3

= 1 x 5 + 6 x 10 + 15 x 10 = 215