permutation and combination help please..?

Question

1) In election 3 persons are to be elected from 6 candidates. A voter can cast any number of votes but not more than candidates to be elected. In how many ways can he cast his vote.

2) There are 10 students in a class in which three A,B,C are girls. The number of ways to arrange them in a row when any 2 girls out of 3 never comes together is.?

3) The number of different 4 letter words that can be formed out letters of the word MORADABAD is.

=> in 3rd question, i know from theory in my book that it can be calculated as coeff. of x^4 in 4!(1+x)^4 (1+x+(x^2/2!)) (1+x+(x^2/2!)+(x^3/3!)) as there are 3'A' , 2'D' and 1 each of M,R,O,B. and it is giving me correct answer but calculating coeff. is taking so much time (and i personally really hate it). I want to know is there any shortcut or a simpler way to solve this question.

thanks in advance.

M3 · Accepted Answer

1/
Each person can either cast/not cast his vote for a person,
so 2^6 = 64, but at least 1 person has to be voted for, so 63 ways

2/
• 1 • 2 • 3 • 4 • 5 • 6 • 7 •
the 7 boys can be permuted in 7! ways,
the girls can be placed in 8P3 = 8*7*6 ways in • spots
thus ans = 7!*8*7*6 = 1693440

3/
3-1 of a kind: 1c1*5c1*4!/3! = 20
2-2 of a kind: 2c2*4!/2!2! = 6
2-1-1 of a kind: 2c1*5c2*4!/2! = 240
1-1-1-1 of a kind: 6c4*4! = 360
adding up, ans = 626 <-------

edit:
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the ans of 626 tallies with the formula
https://www.wolframalpha.com/input/?i=coefficient+of+x%5E4+in+4%21%5B%281%2Bx%29%5E4+*%281%2Bx%2B%28x%5E2%2F2%21%29%29+*%281%2Bx%2B%28x%5E2%2F2%21%29%2B%28x%5E3%2F3%21%29%29%5D+

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permutation and combination help please..?

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