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

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

The fifth position corresponding to............... AAAB could be

AAABB

The fifth position corresponding to............... AABA could be

AABAB

The sixth position corresponding to AAABB- cannot be a B

because it creates a tie so it must be an A to give:

it creates a tie so it must be an A to give:

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.