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I only know how to solve the second one. \[C(10,2)\frac{ 10! }{ 2!(10-2)! }\]
I don't think it is 10C2. I think it is 10P3.
Yes I meant 3. Sorry.
I think it is \[ 10P3=\frac{10!}{(10-3)!}=10\times9\times8=720\\ 10C3=\frac{10!}{3!(10-3)!}=120 \]
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if u can explain the meaning of getting doubles i can answer ur first question
I guess having a double is having two dice with the same value?
so the first answer would be 720 and the second 120?
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