Serial numbers for a product are to be made using 2 letters followed by 2 digits. The letters are to be taken from the first 6 l
etters of the alphabet, with no repeats. The digits are taken from the 10 digits 0, 1, 2, ... , 9 , with no repeats. How many serial numbers can be generated?
Alright so I'm coming up with this on the fly; you have the first six letters (a,b,c,d,e,f) and 0-9 and your ten numbers. calculate the amount of possible combinations for the letters by simply writing them down. ab, ac, ad, ae, a f- five bc, bd , be, bf- four cd, ce, cf- three de, df- two ef,- one. adding these all together gets a total of 15 for the letters. now the numbers 01, 02, 03, 04, 05, 06, 07, 08, 09- nine 12, 13, 14, 15, 16, 17, 18, 19- eight 23, 24, 25, 26, 27, 28, 29- seven 34, 35, 36, 37, 38, 39- six 45, 46, 47, 48, 49- five 56, 57, 58, 59- four 67, 68, 69- three 78, 79- two 89- one added together with a total of 45 combinations. alright so, 45 different number combinations and 15 letter combinations. multiplying 15 by 45 should tell you the total possible combinations for a two letter and two number serial-number
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)
first factor into (x-r1)(x-r2)... form
p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16) group the like ones p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
multiplicity is how many times the root repeats in the function for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2
so
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16) (x-4)^8 is the root 4, it has multiplicity 8 (x-(-4))^1 is the root -4 and has multiplicity 1 (x-1)^1 is the root 1 and has multiplity 1 (x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real
baseically
(x-4)^8 is the root 4, it has multiplicity 8 (x-(-4))^1 is the root -4 and has multiplicity 1 (x-1)^1 is the root 1 and has multiplity 1
