A friend sent me this. I don't hav

[RandyRiegel]

A friend sent me this. I don't have an answer yet but figured I'd post it here:

There 351 tickets in bag.
Each ticket has a letter of A through Z.
There is a different quantity of each ticket as shown ...
A = 1
B = 2
C = 3
...
X = 24
Y = 25
Z = 26

If a person picks out 1 ticket at a time randomly then how many tickets total must be pulled to guarantee that you have 10 Tickets with the same letter?

 

[PHV]

Spoiler

I'd say 198 (1+2+3+4+5+6+7+8+9*18)

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[PHV]

Spoiler

OOps, I meant 199 (1+2+3+4+5+6+7+8+9*18 + 1 to get the 10th)

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[kwbMitel]

[Hide]199[/hide]

**********************************************
What's most important is that you realise ... There is no spoon.

 

[CajunCenturion]

[hide] ==> to guarantee that you have 10 Tickets with the same letter?
The emphasis being on guarantee.

First, you'd have to allow for pulling all the tickets A=1 through I=9, which accounts for 10 * 4.5 = 45 tickets.
Secondly, you'd have to allow for pulling nine tickets from each of the remaining 17 letters = 9 * 17 = 153 tickets
Then you'd have to pull one more to be ensured of having 10 tickets of the same letter.
45 + 153 + 1 = 199
[/hide]



--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
Wise men speak because they have something to say, fools because they have to say something. - Plato

 

[Olaf Doschke]

You shouldn't post and you shouldn't answer math homework questions.

Bye, Olaf.

 

[RandyRiegel]

Olaf,

This is not a homework question. I've been out of college for 16 years, now your making me feel old.

Randy

 

[kwbMitel]

It hadn't occurred to me that it might be homework but the simplicity supports that theory. Glad it wasn't, I hate being duped

**********************************************
What's most important is that you realise ... There is no spoon.