Combinations Question ; I must be missing something

[hms.tech]

Homework Statement

A large Box of biscuits contains nine different varieties. In how many ways can four biscuits be chosen if :
a)Two are the same and the others different
b)Two each of two varieties are selected
c)three are the same and the fourth is different

Homework Equations

Combinations formulas

The Attempt at a Solution

a) The number of Choices for the first one are "9"
for the 2nd one is: "1"
for the third one are: "8"
for the fourth one are "7"

Ans : 504 (correct ans : 252 ) [How ?]

b) The number of choices for the 1st : "9"
2nd : "1"
3rd : "8"
4th : "1"
Ans : 72 (correct ans : 36) [How ?]

c) 1st choice : "9"
2nd choice : "1"
3rd Choice : "1"
4th choice : "8"
Ans : 72 (correct ans : 72) [Hmmm...]

 

[Fightfish]

a) Your method results in double-counting.
Essentially your method is:
(1) Pick 1 variety out of the 9. Let's have two biscuits of this.
(2) Pick 1 variety out of the 8 varieties left. One biscuit.
(3) Pick 1 variety out of the 7 varieties left. One biscuit.

Steps (2) and (3) double count, because your method is assuming that the order matters, which does not. eg. picking biscuit A in step 2 and B in step 3 is the same case as picking biscuit B in step 2 and A in step 3.

b) Same problem of double counting.

c) No double counting because of asymmetry between the cases (i find myself tempted to say that symmetry is broken between the two distinct varieties of biscuits but anyway :p). ie three biscuit A + one biscuit B is not the same as three biscuit B + one biscuit A. That's why you got the correct answer.

 

[hms.tech]

a) Your method results in double-counting.
Essentially your method is:
(1) Pick 1 variety out of the 9. Let's have two biscuits of this.
(2) Pick 1 variety out of the 8 varieties left. One biscuit.
(3) Pick 1 variety out of the 7 varieties left. One biscuit.

Steps (2) and (3) double count, because your method is assuming that the order matters, which does not. eg. picking biscuit A in step 2 and B in step 3 is the same case as picking biscuit B in step 2 and A in step 3.

b) Same problem of double counting.

c) No double counting because of asymmetry between the cases (i find myself tempted to say that symmetry is broken between the two distinct varieties of biscuits but anyway :p). ie three biscuit A + one biscuit B is not the same as three biscuit B + one biscuit A. That's why you got the correct answer.

You are right, I must divide them by "two" . I understand !