Solution to review question 1.1.6A. (i) 3!= 3 ( 3 − 1 )( 3 − 2 )= 3 × 2 × 1 = 6
(ii) 6!= 6 × 5 × 4 × 3 × 2 × 1 = 720(iii)24!
23!=24 ×23!
23!= 24(iv)12!
9!3!=12 × 11 × 10 ×9!
9!3!=12 × 11 × 10
6= 220B. (i) (a)^3 C 2 =3!
( 3 − 2 )!2!=3!
1!2!= 3(b)^6 C 4 =6!
( 6 − 4 )!4!=6 × 5 ×4!
2!4!= 15(c)^6 P 3 =6!
3!= 6 × 5 × 4 = 120(ii) Two letters can be chosen fromABCDin(^4) C
2 =
4!
2!2!
=6ways
(iii) There are 5!=120 permutations of five different letters.
1.2.7 Powers and indices
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429 ➤
Powers, or indices, provide, in the first instance, a shorthand notation for multiplying a
number by itself a given number of times:
2 × 2 = 22
2 × 2 × 2 = 23
2 × 2 × 2 × 2 = 24
etc.
For a given numberawe have
an=a︸×a×a︷︷×...×a︸
ntimes
ais called thebase,nthepowerorindex.a^1 is simplya. By convention we takea^0 = 1
(a=0). We introducea−^1 to denote thereciprocal
1
a
, since then 1=a×
1
a
=a^1 ×
1
a
a^1 ×a−^1 =a^1 −^1 =a^0 follows. In general,a−n=
1
an
. From these definitions we can derive
therules of indices:
am×an=am+n