Question

A school is using identification cards (ID cards) that consist of 3 letters selected from A to E inclusive followed by 3 digits chosen from 0 to 9 inclusive.
a) How many different ID cards can be issued to students if a digit may be used more than once but all 3 letters of each ID are different?

b) New ID cards are issued to all students each year and the old cards discarded. However, the old ID numbers are not used again. If, on average, the school's population increases by 10% each year and was 2000 during the year when the ID cards were first used, how many years will elapse before cards with numbers already used will have to be issued?

Answer 1

I've done part A and need help with part B now.

Answer 2

and I believe part B relates to creating the right equation except that I couldn't.

Answer 3

Is the answer 216??

I am not sure??

Answer 4

nope, it's the 15th year.

Answer 5

Wait..

Answer 6

in the 15th year*

Answer 7

How many total numbers of ID cards can we make do you know that??

Answer 8

yep, 60000.

Answer 9

You are giving the answer for a part right??

Answer 10

for part b I tried creating an equation of 2000+200n=60000 but it's not correct.

Answer 11

yep

Answer 12

I got 2000 + 200x = 43200


Ha ha ha..

Wait let check it once more..

Answer 13

yeh neither urs or mine is correct.

Answer 14

On right hand side we must get 5000:

2000 + 200x = 5000
200(x+10) = 5000

x = 15

But I am not getting 5000..

Answer 15

why 5000? and not 60000?

Answer 16

I think it is asking for numbers only..

Read the b part carefully..

Answer 17

See, you have found 60000 for first part and that is right but in first part it is not the answer you will apply to all other parts..

First part is given with some condition and you have found 60000 according to that condition..

Getting??

Answer 18

I got the solution...

Answer 19

but shouldn't we use 60000?

Answer 20

No we have to find it again according to the B part sayings..

Answer 21

since that is the total number of codes that can be produced.

Answer 22

Read the b part carefully dear..

Answer 23

I've read it, I still don't get it.

Answer 24

Read the last line:

before cards with numbers already used will have to be issued?

cards with numbers and not alphabet

Right??

Answer 25

Cards are containing three alphabets and three numbers B part is demanding for card with numbers and not card with alphabets..

Getting it or not??

Answer 26

yepyep.

Answer 27

Now,

find the total numbers of cards now..

See, we can choose 3 numbers from 0 to 9 in 10*10*10 ways right???

Answer 28

yep.

Answer 29

We get 1000..

But in front there are alphabet also..

So there must be alphabets in front but we need not to arrange them here..

For arranging them you will get 60 as your answer..

But here arranging does not matter we are dealing with numbers here...

So, out of 5 how many ways you can use letters in front??

I think 5 ways..

Because you can have A in first , B or C or D or E so total 5 ways..

Getting or not??

Answer 30

So total cards = 5*1000 = 5000


after this you will get your answer:

2000 + 200n = 5000

get n from here..

Answer 31

Able??

Answer 32

I don't get ur
"But here arranging does not matter we are dealing with numbers here...

So, out of 5 how many ways you can use letters in front??

I think 5 ways..

Because you can have A in first , B or C or D or E so total 5 ways."

Answer 33

I understand how there is 60 ways for the three spots from 5x4x3=60 and that there are 5 ways for the first slot.

Answer 34

and why do you times five?

Answer 35

to get 5000?

Answer 36

Because the Cards have formed from 3 alphabets \(\color{green} {AND}\) 3 numbers..

And means multiply...

Answer 37

I will have to go now..

I will surely explain you when I will come online..

Don't mind..

Answer 38

yeh, i have to go as well, I think I will just ask my teacher the day after.

Answer 39

but thanks for your help.

Answer 40

Welcome..

But I couldn't lead you to your answer..

Sorry..

Answer 41

it's fine.