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1. 2*(263*103) [5.1.18]

2. You can choose 21*21 integer pairs. In the given range 11 are even, 10 are odd. So: (11*11 + 10*10)/21*21 [5.1.20]

3. Count the numbers from 00000 to 99999 - how many 5's appear? C(n,1)*94+C(n,2)*93+C(n,3)*92+ C(n,4)*91 + 1. [5.1.30]

4.(n+1)4 - 1 -- you can pick 0 .. n of any letter, but you can't pick all 0s (hence the - 1 term)

5. 7! / 2!

6. C(26,6)

7. (13!)4 / ((52!)/(4!)13)

8. 20*19*18...*11

9. 4! * (5*6*7)

10.C(11,6)

11. 2n-1

12. 11!/(4!4!2!)

13. (5!/2!)*(2*6*5 + 6*5)

14. C(9,7)*C(8,6)*C(9,7); C(9,7)*C8,6)*C(6,4)

15. C(103,99)

16. Imagine n groups of n independently identical objects. By the "bananas" principle, this can be arranged in (n2)!/(n!)n distinct ways.

But now if you declare that two arrangements are identical if (the n) independently identical groups are rearranged as well, then you've "added" another group of n! to the denominator.

-- RobbieMoll - 2012-04-26

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Topic revision: r6 - 2012-04-30 - RobbieMoll

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