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A friend sent me this. I don't hav

A friend sent me this. I don't hav

(OP)
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?

RE: A friend sent me this. I don't hav

Hidden:

199

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

RE: A friend sent me this. I don't hav

Hidden:

==> 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

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RE: A friend sent me this. I don't hav

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

Bye, Olaf.

RE: A friend sent me this. I don't hav

(OP)
Olaf,

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

Randy

RE: A friend sent me this. I don't hav

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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