Fairly Simple Counting Questions

[knight247]

1) If the letter of the word MANAGEMENT are arranged in different orders, in how many of the arrangements will the vowels be in even positions?
(A)2880
(B)5400
(C)17280
(D)21600
(E)86400

2)In how many ways can a group of 4 persons be seated on 5 chairs placed around a circular table?

[shankar.ashwin]

MANAGEMENT - 10 Letters ( 2A / 2E) (2M/2N/G/T)

5 even positions for 4 vowels. Picking positions for vowels - 5C4 = 5

Arrangement of vowels - 4!/2!2! = 6
Arrangement of rest - 6!/2!2! = 180

Together = 5*6*180 = 5400.

B IMO

[knight247]

The consonants from the word Management are
M_N_G_M_N_T_
The four vowels can be arranged in any of the 6 gaps above. Shouldn't it be 6C4 instead of 5C4?

[shankar.ashwin]

Management has 10 letters, why are there 12 slots? Am I wrong here?

The consonants from the word Management are
M_N_G_M_N_T_
The four vowels can be arranged in any of the 6 gaps above. Shouldn't it be 6C4 instead of 5C4?

[knight247]

Management has 10 letters. But don't we need to consider the scenario where a vowel would be in the last spot. Example MNGAMANETE?? Would such a scenario be covered in 5C4?

[baladon99]

1) There are 10 letters in MANAGEMENT . therefore 5 even positions.

available vowels - A , A ,E , E .
A can be placed in any of the 5 positions.
2nd A in any of the 4 positions.
E- 3 positions.
2nd E - 2 positions. So totally - 5*4*3*2 = 120 positions out of which there are 2 repetitions .
So 120/2!2! = 30 positions.

There are 6 consonants which can be arranged in 720 ways.
Out of which there are two repetitions M and N .
So number of ways = 720/2!2!= 180

Total number of arrangements = 180*30 =5400

2) I think that it is 4! ways as there would be an empty chair in-between.

[shankar.ashwin]

Think of it this way I have 5 slots and 4 Nos

_ _ _ _ _ and 1,2,3,4.(The last slot corresponds to the 10th in this problem)

We can pick 5 possible slots for these 4 in 5C4 ways (If there were only 4, we have only 1 option of filling out the 4 slots with 4 vowels)

We multiply by 5 to consider those cases as well. (if we didnt consider the 10th slot, we would be left with only 4 slots and 4 vowels)

Management has 10 letters. But don't we need to consider the scenario where a vowel would be in the last spot. Example MNGAMANETE?? Would such a scenario be covered in 5C4?

[sl750]

M_N_G_M_N_T_

Number of ways of arranging the consonants is 6! ways
We have 6 slots and 2 vowels. We can select any six slots for the first vowel followed by any 5 slots for the other vowel. This gives us 30 such combinations

Total = 6!*30 = 21600

[shankar.ashwin]

2)In how many ways can a group of 4 persons be seated on 5 chairs placed around a circular table?

4 people can choose from 5 chairs in 5C4 ways - 5
Circular arrangement for 5 chairs - (5-1)! - 4! ways.

Totally, 5*4! = 5! = 120 ways IMO.

[GmatMathPro]

I agree with 5400 for the first question. Setting it up as M_N_G_M_N_T_ doesn't work because you will end up with gaps in the word, which is not allowed, because MANAGEMENT has 10 letters, not 12. for example, let's say you pick the 4th, 6th, 8th, and 10th spots for the vowels and you have: M_NAGAMENET_. So what do you do about those gaps? well, maybe you argue that all we're doing is determining the possible sequences of the letters so we can just remove the gaps. But if you remove the gaps, you get MNAGAMENET. But now you're totally screwed because without the gap, the vowels have shifted back to odd positions.

I did it the same way as baldon99, but shankar's way looks fine too.

2) How many ways can 4 people be seated in 5 chairs around a circular table?

Two ideas:
1. Pretend there is a ghost. If we're seating 4 people and a ghost, that is the same as seating 5 people, with the ghost representing the empty chair. Then it's just 5!/5=4!.

2. Pretend there are 4 chairs and 4 people. This is obviously 4!/4=3!, but now we have to pick a spot for the empty chair to go. For any of the 3! unique arrangements, it can go between persons 1 and 2, 2 and 3, 3 and 4, or 4 and 1. So we have 4 choices for where to put the empty chair. 3!*4=4!

[shankar.ashwin]

I was thinking the same too, 1 empty chair is similar to one unique person. 4! ways.

5! ways would be if there were 6 people and 6 chairs, here we have only 5 chairs and 4 people, so it should definitely be a much smaller number.

[n@resh]

2)In how many ways can a group of 4 persons be seated on 5 chairs placed around a circular table?

Answer: 5P4/4 = 30 ways?