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
Answer:
what are the answer choicesss
Step-by-step explanation:
Yes, actually a couple:
different number with same digits (ex):
7.05
7.50
5.07
5.70
0.57
0.75
705
70.5
75
750
570
etc
hope this helps
Bear in mind that multiplying anything, and anything whatsoever by 1, has a product of the anything itself.
so, let's multiply 2/3 by 1 then and check about,
