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A friend sent me this. I don't hav

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RandyRiegel

Programmer
Sep 29, 2003
256
US
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?
 
I'd say 198 (1+2+3+4+5+6+7+8+9*18)

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886
 
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
 
[Hide]199[/hide]

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



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

Bye, Olaf.
 
Olaf,

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

Randy
 
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.
 
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