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1. ### What four-letter sets are possible using the letters A, B, C, and D?

AAAA AAAB AAAC AAAD AABA AABB AABC AABD AACA AACB AACC...

3 Answers · Science & Mathematics · 09/10/2010

2. ### Probability problem - Urgent help needed!?

... BBAA BBBB AAAB ABAB ABBB AABA ABBA BABB ABAA BAAB BBAB BAAA...

2 Answers · Science & Mathematics · 26/06/2010

3. ### given word BANANA, how many distinct 4- letter words can be formed using just the A's and 1 other letter?

BAAA ABAA AABA AAAB NAAA ANAA AANA AAAN There's your 8.

1 Answers · Science & Mathematics · 04/03/2010

4. ### Difficult Maths Question?

...10 choices for A, 9 choices for B = 90 ways AABA --> 10 choices for A, 9 choices for B = 90 ...

1 Answers · Science & Mathematics · 24/04/2009

5. ### Two questions on partial orderings?

...e = 5) eaaa eaab eaac eaba eabe eaca caaa caab caac caba cabe caca aaaa aaab aaac aaba aabe aaca abaa abab abac abea acaa acab acac baaa baab baac baba babe baca beaa beab...

2 Answers · Science & Mathematics · 09/03/2009

6. ### Group Theory Proof -- Simple?

... aba⁻¹ = b⁻¹ by a from left, by a⁻¹ from right: aaba⁻¹a⁻¹ = ab⁻¹a⁻¹, but the right side here ...

3 Answers · Science & Mathematics · 22/02/2009

7. ### Any simpler method to expand (a+b+c+d)^4 ?

...product has a similar big expansion, only much longer: aaaa+aaab+aaac+aaad+aaba+aabb+aabc+aabd+aaca+aacb+... If you think about it for a while, you could certainly...

2 Answers · Science & Mathematics · 22/08/2008

8. ### Serious math: find a formula for the number of surjective function from a set with n elemente to a set with m?

...we can get 3^4 possible outcomes: AAAA AAAB AAAC AABA AABB AABC AACA AACB AACC ....etc. ...

3 Answers · Science & Mathematics · 03/07/2008

9. ### Probability problem?

The items could sell in various orders: AABA ... BAAB ... BBAA ... etc. the one we are interested...

1 Answers · Science & Mathematics · 11/06/2008

10. ### the game RISK and its card trading's probability?

You don't need to ignore the wild cards, and you don't calculate it the way you were doing. Suppose there are 14 cards of each type (soldier, horse, cannon) and 2 wild cards. The only way to need five cards is to be unable to trade...

5 Answers · Science & Mathematics · 17/11/2007