nverma wrote:
What is the number of different ways to choose a chairman, two deputies, and two assistants for the class committee out of 7 students up for elections.
A. \(7C1*6C1*5C1*4C1*3C1\)
B. \(7C1*6C2*4C2\)
Which solution is correct ? And Why ?
Correct answer is B: \(C^1_7*C^2_6*C^2_4=630\).
\(C^1_7\) - # of ways to choose 1 chairmen out of 7;
\(C^2_6\) - # of ways to choose 2 deputies out of 6 members left;
\(C^2_4\) - # of ways to choose 2 assistants out of 4 members left.
Answer A is not correct because it's counting # of different ways to choose 2 deputies out of 6 as \(C^1_6*C^1_5\) (and next # of ways to choose 2 assistants out of 4 as \(C^1_4*C^1_3\)) which is not right. \(C^1_6*C^1_5\) will have duplications in it and needs to be divided by 2! (# of peoples), which then gives the same answer as \(C^2_6\).
Consider this: in how many different ways we can choose 2 different letters out of A, B, and C?
AB
AC
BC
Only 3, which can be obtained by \(C^2_3=3\), another way would give incorrect answer - \(C^1_3*C^1_2=6\).
The original question can be solved in another way:Members: 1 - 2 - 3 - 4 - 5 - 6 - 7. Positions: C (chairmen), D (deputy), D (deputy), A (assistant), A (assistant), N (no position), N (no position): CDDAANN. # of ways to assign each letter (each position) to the members would be the # of permutations of 7 letters CDDAANN = \(\frac{7!}{2!2!2!}=630\).
1-2-3-4-5-6-7C-D-D-A-A-N-N
D-C-D-A-A-N-N
D-D-C-A-A-N-N
...
...
Hope it helps.