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Senior one won the holiday math problem.
I'm here to get points. They are all "original". I'm exhausted. Give them to me.

1. Divide into three piles first: only 122 C52*C32, and then allocate A33.

180

2. (People are different) * * * 4 6 species, piling up 0 123 C63*C32, with only one distribution mode.

10/2 16

3. (People are considered to be different) People with strong C2N are in different groups C(2n-2)(n- 1).

3n-2/2(2n- 1)

4. Just consider one.

I calculated 32/8 1.

5. (Binding and sewing, different people) * * ***A 10 10/0/0, Martians bound the whole earth A66, Martians all arranged A33 and Mercury inserted A72.

I also worked out 1/20.

6. very troublesome. What I did at that time was to remember the answer to this question.

7. Turning to the arrangement problem, 15 balls are all arranged. A 15 15 requires the fifth ball to be red, and A 14 14.

1/ 15

8. Of course it's unfair. Superman won 0.6+0.12 * 0.6+0.12 * 0.12 * 0.6. ....

Forget it.

9.P (non-B/A)= 1 means that non-B/A must occur. If A is not B, if P(AB) is not equal to 0, that is, AB can occur at the same time. If A can also be B, this is contradictory.