permutations and combinations ? (1 Viewer)

[jnney]

Find the number of arrangements of the letters in the word PENCILS if E precedes I.

In how many ways can 3 boys and 2 girls be arranged in a row if a selection is made from 5 boys and 4 girls? (7200) In how many of these arrangements does a boy occupy the middle position?

 

[Aysce]

What is the answer for the first one? I might have it

 

[jnney]

2520

 

[deterministic]

For the first one, there are 2 cases:
(i)I precedes E
(ii) E precedes I.

By symmetry, there should be exactly the same number of arrangements for (i) and (ii). Hence:
number of arrangements = total arrangements/2 = 7!/2=2520

 

[Shadowdude]

For the first one, I did it the 'long' way so you can see how it works. I cut it into cases.

1. Number of arrangements of the letters in the word 'PENCILS' if E comes before I.

Cases: "EI", "E_I", "E__I", "E___I", "E____I", "E_____I", "E_____I"

1: "EI".

1. Consider "EI" as one letter, so there are now six letters to arrange: P, N, C, L, S, EI. Place "EI" in the word... 6 ways
2. Place the other 5 letters in the arrangement of the word... 5! ways

Sub-total: 6*(5!) = 720

2. "E_I"

Consider "E_I" as one letter, so there are five letters to arrange.

1. Select the letter missing in "E_I", choose from, P, N, C, L or S. ... 5 ways
2. Place "E_I" in the arrangement... 5 ways
3. Consider "E_I" as one letter, so there are four letters to arrange... 4! ways

Sub-total: 5*5*(4!) = 600

3. "E__I"

Similarly:

1. Select the two letters missing in "E__I", choose from: P, N, C, L or S ... C(5,2) ways ("5C2")
2. Arrange the two letters missing... 2! ways (as they are different)
3. Place "E__I" in the arrangement... 4 ways
4. Consider "E__I" as one letter, and there are three letters to arrange... 3! ways

Sub-total: C(5,2)*2!*4*3! = 480

4. "E___I"

Similarly:

1. Select three letters for the missing ones... C(5,3) ways
2. Arrange them within the 'letter' "E___I"... 3! ways
3. Place "E___I" in the arrangement... 3 ways
4. Consider "E___I" as one letter, there are two letters to arrange... 2! ways

Sub-total: C(5,3)*3!*3*2! = 360

5. "E____I"

Similarly:

1. Select four letters for the missing ones in "E____I"... C(5,4) ways (alternatively: pick the letter that does not show up)
2. Arrange them within the 'letter' "E____I"... 4! ways
3. Place "E____I" in the arrangement... 2 ways
4. Place the other letter... 1 way

Sub-total: C(5,4)*4!*2*1 = 240

6. "E_____I"

1. Select the five letters missing in "E_____I"... 1 way
2. Arrange them within the letter... 5! ways

This 'letter' "E_____I" is made up of seven letters and thus is an arrangement in itself.

Sub-total: 5! = 120

TOTAL: 720 + 600 + 480 + 360 + 240 + 120 = 2520

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Also what's the second question word for word, and what's the 7200 in brackets? It's a bit confusing.

 

[Sy123]

Find the number of arrangements of the letters in the word PENCILS if E precedes I.

In how many ways can 3 boys and 2 girls be arranged in a row if a selection is made from 5 boys and 4 girls? (7200) In how many of these arrangements does a boy occupy the middle position?

1) In this case there are only two types of events, it is either E is before I or E is AFTER I
Because E and I cannot be in the same slot, therefore of the total possibilities only half those events are E before I

PENCILS

7 letters, so total possiblities is

7!=5040

5040/2 = 2520 (by the way, I got an extremely similar question in a quiz we got, no one got it right, so the teacher at the end explained it)

2) Of the 5 boys, you select 3 boys, and of the 4 girls you select 2
So total possibilties

5C3 x 4C2 = 60 possiblities

2i)Let Boys=B Girls=G

One boy has been sacrificed to be in the middle of the row, so he is forgotten among our selections
So we select 2 boys from a group of 4
4C2 x 4C2 =36?



Im not sure about the last answer, it is probably wrong

 

[Shadowdude]

I'm also getting some weird answers for part 2 of the second question... that's why I want to know what the question is in full.

First part's 7200 seems to be:

1. Pick 3 boys from the 5... C(5,3)
2. Pick 2 girls from the 4... C(4,2)
3. Arrange the five people in a row... 5! ways

So: C(5,3)*C(4,2)*5! = 7200.

Second part... still trying to figure it out.

EDIT: I think I got it, hold on...

 

[Shadowdude]

Let B = boys, G = girls

A required arrangement is: __B__ , so the boy is in the middle. There are three boys and two girls to arrange. So:

1. Pick the boy who goes in the middle... 3 ways
2. Arrange the other 4 chosen people around him... 4! ways

Sub-total: 3(4!) = 72.

Now, the boys and girls are not set. They can change. In fact, there are C(5,3) = 10 ways to pick the boys and C(4,2) = 6 ways to pick the girls to be arranged.

We multiply the sub-total by 60. This is because the 'boy' in the middle can be one of five boys, and the other two boys to be arranged in a row can also be one of five boys... and similarly, for the girls.

Essentially, given a set group of 3 boys and 2 girls - there are 72 ways to arrange them in a row where a boy is in the middle. There are 60 ways to select the group of 3 boys and 2 girls. Thus...

Answer: 72*60 = 4320.


Do I win? =P