There are 4 A's and 3 B's. If all these seven items are distinct then there are 7!
different ways to get them one at a time without replacement. Since the four A's
are indistinguishable and so are the three B's, there are 7!/ (4! 3!) different ways
in which we can draw the 4 A's and 3 B's one at a time without replacement.

OR

Out of the seven positions choose three for the B's.

NOTATION please note that nCr means combination of r items taken from n items.

7C3

A definite way to list all these outcome:

Possible positions of the B's are 1,2,3,4,5,6,7.

Since out of 7 votes three are B's and the remaining are A's; the possible positions of
the three B's and the corresponding outcomes are :

123 124 125 126 127   BBBAAAA    BBABAAA     BBAABAA    BBAAABA    BBAAAAB
134 135 136 137          BABBAAA     BABABAA    BABAABA    BABAAAB
145 146 147                  BAABBAA     BAABABA    BAABAAB
156 157                         BAAABBA     BAAABAB
167                                BAAAABB
These are  6C2  = 15   outcomes

234 235 236 237      ABBBAAA    ABBABAA     ABBAABA    ABBAAAB
245 246 247              ABABBAA     ABABABA    ABABAAB
256 257                     ABAABBA     ABAABAB
267                            ABAAABB
These are 5C2 = 10 outcomes.

345 346 347        AABBBAA    AABBABA     AABBAAB
356 357                AABAABA     AABABAB
367                       AABAABB
These are 4C2 = 6 outcomes.

456 457            AAABBBA AAABBAB
467                    AAABABB
These are 3C2= 3 outcomes.

567                       AAAABBB
This is  2C2 = 1 outcome. Note that 15 + 10 + 6 + 3 + 1 = 35.

(b)  In order that A should be ahead at all seven draws one at a time.
The first two draws must be AA because otherwise either B will be
ahead or there is a tie. Similarly the seventh vote can not be an A because
in the remaining six draws there are three A's and three B's and hence there
will be a tie at some stage, within the first six draws. This gives the structure
AA - - - - B, for A to stay ahead.
From here on there are two ways of doing this list of outcomes:

Method I
The third position could be:
AAA
AAB
The fourth position corresponding to AAA-
could be:
AAAA
AAAB
The fourth position corresponding to AAB-
cannot be a B because then you have a tie. Therefore it should be
AABA
The fifth and sixth position corresponding to  AAAA-- should be
AAAABBB
The fifth position corresponding to...............  AAAB could be
AAABABB
AAABB
The fifth position corresponding to...............  AABA could be
AABAABB
AABAB
The sixth position corresponding to AAABB- cannot be a B
because it creates a tie so it must be an A to give:
AAABBAB
The sixth position corresponding to AABAB- cannot be a B because
it creates a tie so it must be an A to give:
AABABAB.
All cases have been exhausted because seventh position must be a B.
The answer has the above five outcomes in bold and italics.

Alternately you could go:

The four empty positions have two A's and two B's, which can be filled
4C2= 6 ways: BBAA, BABA, BAAB, ABBA, ABAB, AABB.

Out of these possibilities the first one namely
AABBAAB does not work because of a tie at position four.

Hence the final answer is AABABAB, AABAABB, AAABBAB,
AAABABB, AAAABBB.