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.
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.
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
These are 6C2 = 15 outcomes
234 235 236 237 ABBBAAA ABBABAA
245 246 247 ABABBAA ABABABA ABABAAB
256 257 ABAABBA ABAABAB
These are 5C2 = 10 outcomes.
345 346 347 AABBBAA AABBABA
356 357 AABAABA AABABAB
These are 4C2 = 6 outcomes.
456 457 AAABBBA
These are 3C2= 3 outcomes.
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:
The third position could be:
The fourth position corresponding to AAA-
The fourth position corresponding to AAB-
cannot be a B because then you have a tie. Therefore it should be
The fifth and sixth position corresponding to AAAA-- should be
The fifth position corresponding to............... AAAB could be
The fifth position corresponding to............... AABA could be
The sixth position corresponding to AAABB- cannot be a B
because it creates a tie so it must be an A to give:
The sixth position corresponding to AABAB- cannot be a B because
it creates a tie so it must be an A to give:
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,