I need help regarding perms and comb questions: a few answers would help me to gain the momentum*?

*
1) a girl wishes to phone a friend but cant remember the exact number. She knows that it is a 5-digit number, i.e. even, and it consists of the digits 2,3,4,5 and 6 in some order. Using this info, find the largest number of different wrong telephone numbers she could try.

2) In how many ways can a committee of 3 men and 3 women be chosen from a group of 7 men and 6 women? I know this is 7C3 and 6C3 but look at the 2nd part:- 'the oldest of the 7 men is A and the oldest of the 6 women is B. It is decided that the committee can include at most one of A and B. In how many ways can the committee now be chosen?

3) A 10-digit number is formed by writing down the digits 0,1,2,3,4,5,6,7,8,9 in some order. No number is allowed to start with 0. Find how many such numbers are odd.

4) a)i) 8 ppl go to the theatre and sit in a particular group of 8 adjacent reserved seats in the front row. 3 of the 8 belong to 1 family and sit toghter. If the other 5 ppl dont mind where they sit, find da arng

4)a)ii) if the other 5 ppl dont mind where they sit, except the two of them refuse to sit togather, find the number of possible seating arrangements for all 8 ppl. (tough)

5) in this question, a word is defined to by any set of letters in a row, whether or not it makes sense. Find how many different words can be made using all 8 letters of the word SYLLABUS.

1) Since the digits 2, 3, 4, 5 and 6 are all different, there are a
total of 5! = 120 possible permutations (i.e. phone numbers ). Therefore, she has to dial 119 wrong numbers in the worst case.

EDIT: Hmm... It seems you said "even". If it's even, then it ends in either 2, 4 or 6. Half of those 120 permutations are even. The answer then becomes 59 wrong numbers.

4) a) i)

3! for permuting the members of the family
5! for permuting the other members
C(8,3) for the number of places the 3 members of the family, as a group, can sit on.

3! * 5! * C(8,3)

5) The word is "SYLLABUS". There are two instances of the letter L, and two instances of the letter S. Therefore, we just have to divide all possible permutations by 2! (two L's) times 2! (two S's).

number_of_words = 8! / 2! / 2! = 8! / (2! * 2!) = 8! / 4 = 8*7*6*5*4*3! / 4 = 8*7*6*5*3*2*1 = 10080