1. ## Probability

I have got stuck on this question, is anyone able to help me please (any help is welcome):

Given an alphabet of 26 letters, how many sets of initials can be formed if every person has one surname and

i) exactly 2 given names

ii) at most 2 given names

iii) at most 3 given names

iv) The conclusion is that in a town of 20000 inhabitants then either some people have the same set of initials or at least x have more than 3 initials. How large is x?

Many thanks to anyone who has any advice.

2. Here is what I'd do.....

(i) There are 26 possible letters, each set of initials would have 3 letters in it, order doesnt matter, i.e. ABC does not equal ACB. So type in 26P3 on your calculator (P=permutation).

(ii) This time each set of initials can have 2 or 3 values, so calculate 26P3 + 26P2.

(iii) 26P4 + 26P3 + 26P2.

(iv) x = solution for (ii) - 20,000

Hope this helps...

3. roegamma,

thanks for ur help, by permutation (P) do u mean type 26^3 into the calculator.

4. No, you should have a P button somewhere on your calculator. Could be signified by nPr.

http://www.maplin.co.uk/images/Full/A92FQ.jpg

Thats pretty much my calculator. For me to do 26P4 I'd have to type 26, shift, x, 4.

5. roegamma

oh yeah i found it thanks. one last question, for part (iv) should it be 20000 - (26P3 + 26P2) otherwise if i did (26P3 + 26P2) - 20000 then i would get a negative number?

6. Yeah, sorry thats my fault. It should be 2000 - answer to part (ii).

7. Originally Posted by smiler
Given an alphabet of 26 letters, how many sets of initials can be formed if every person has one surname and
i) exactly 2 given names
ii) at most 2 given names
iii) at most 3 given names
iv) The conclusion is that in a town of 20000 inhabitants then either some people have the same set of initials or at least x have more than 3 initials. How large is x?
There is nothing about the statement of this problem to lead one to conclude that a person has three different initials. I know someone named August Aaron Adams. His initials are AAA.

Originally Posted by roe_gamma
Here is what I'd do.....
(i) There are 26 possible letters, each set of initials would have 3 letters in it, order doesnt matter, i.e. ABC does not equal ACB. So type in 26P3 on your calculator (P=permutation).
(ii) This time each set of initials can have 2 or 3 values, so calculate 26P3 + 26P2.
(iii) 26P4 + 26P3 + 26P2.
(iv) x = solution for (ii) - 20,000
The above answers assume that the initials are all distinct in a name. That is an unwarranted conclusion.
i) $\displaystyle {26}^3$.

ii) $\displaystyle 26 + {26}^2+{26}^3$ [no given name, one given or two] (Now you may want to assume that everyone has at least one given name.)

iii) is like #2 $\displaystyle 26 + {26}^2+{26}^3 + {26}^4$

To see (iv) note that from (iii) $\displaystyle 26 + {26}^2+{26}^3 = 18278$ (or if everyone must have a given name then $\displaystyle {26}^2 + {26}^3 = 18252$). So in any case in a town of 20000 by the pigeonhole principle some people must have the same initials.

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