Page 34 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series
In problems on permutations and combinations, the number of combinations of choosing r objects from n
n
different objects is given by 1 2 . For example, the number of ways of choosing 3 books out of 10 different
r
10
books is 1 2 , where
3
10
10!
1 2 = 3!7! = 120 ways
3
n
The notation 1 2 is called the binomial coefficient as it is found in binomial expansions.
r
n
The binomial coefficient 1 2 is defined as
r
n
n!
1 2 = (n – r)! r! for n, r  Z and 0  r  n.
+
r
Notice that 2
n
n!
1 2 = (n – 0)! 0! = 1 since 0! = 1
0
n
n!
1 2 = (n – 1)! 1! = n(n – 1)! = n
1
(n – 1)!
n
n!
1 2 = (n – 2)! 2! = n(n – 1)(n – 2)! = n(n – 1)
2
2!
(n – 2)! 2!
n
n!
1 2 = (n – 3)! 3! = n(n – 1)(n – 2)(n – 3)! = n(n – 1)(n – 2)
3!
(n – 3)! 3!
3
So, in general,
n
n!
1 2 = (n – r)! r! = n(n – 1)(n – 2)(n – 3) … (n – r + 1) , r  n.
r
r!
Example 32
Show that
n
n
n
n
=
(a) 1 2 1 n – r 2 (b) 1 2 1 r + 1 2 1 n + 1 2
+
=
r
r
r + 1
Solution: (a) From the definition,
n
n!
1 2 = (n – r)! r!
r
n
n!
and 1 n – r 2 = [n – (n – r)]! (n – r)!
= n!
r! (n – r)!
= n!
(n – r)! r!
n
n
Hence, 1 2 = 1 n – r 2 .
r
(b) From the definition,
n
n
n!
n!
+
1 2 1 r + 1 2 = (n – r)! r! + [n – (r + 1)]! (r + 1)!
r
= n! + n!
(n – r)(n – r – 1)! r! (n – r – 1)! (r + 1) r!
= n! 3 1 + 1 4
(n – r – 1)! r! n – r r + 1
121
02 STPM Math T T1.indd 121 3/28/18 4:21 PM