Permutations question! (1 Viewer)

[MONKEYjulz]

Hey guys,
This is a particular permutation question which annoys me. I use the Cambridge method and get a total different answer to what another textbooks says, cause they have some other way of working it out. I think it was maths in focus, but don't take my word there.
So this ISN'T the exact question out of there. My friend gave it to me

Q: 5 letter codes are formed from the word ALGEBRA, how many codes can be made?

I got 1260, but i think the answer was 1860 or something near there.

What is the "correct" answer and what is your method of working?
Cheers!!

 

[Top Secret]

Should be 1260.

 

[LordPc]

I got 1260, but i think the answer was 1860 or something near there.

What is the "correct" answer and what is your method of working?
Cheers!!

7 letters, pick 5, order important = 7P5

then remove identical cases due to there being 2 'A's

so its 7P5/2! which is 1260

 

[Timothy.Siu]

2C1x5C4x5!+5C3x5!/2! +5!

 

[MONKEYjulz]

2C1x5C4x5!+5C3x5!/2! +5!

see what the hell he got 1920 :S

 

[gurmies]

Not as simple as you guys have done. I think Tim is correct - I got the same answer. You have to take cases where A's appear/don't appear.

 

[MONKEYjulz]

For that same question. How do you work this:

5 letter codes that does not begin or start with A.
Answer: 1860
I got 600

 

[gurmies]

5! + 5C4.2C1.4.4! + (5C3.3.4!)/2! = 1440

Not sure where my logic is wrong...

 

[Timothy.Siu]

For that same question. How do you work this:

5 letter codes that does not begin or start with A.
Answer: 1860
I got 600

5C1x4!+5C1x2C1x4C3x4!+5C1x4C2x4!/2!=13XX something

 

[Drongoski]

Probably going to make a fool of myself again!

Were all 7 letters distinct, we'd have got 7 x 6 x 5 x 4 x 3 = 2520 codes. But because we have 2 'A's in "ALGEBRA" we have:

with 1 'A': 6 x 5 x 4 x 3 x 2 = 720 codes (no problems here)

with no 'A': 5! = 120 codes (again no problems here)

Therefore no of codes with 2 'A's = 2520 - 720 - 120 = 1680 out of which 1680/2! = 840 are distinct.

Therefore total number of codes = 840 + 720 + 120 = 1680 (if that is the correct answer)

 

[Drongoski]

For that same question. How do you work this:

5 letter codes that does not begin or start with A.
Answer: 1860
I got 600

Are you sure book's answer is 1860 and not 1680 ?? Which book is this question from?

 

[k02033]

instead of picking which 5 letters to use, start with picking which 2 letters we are not using. And label the As ie first A is A1 and 2nd is A2

Lets remove 2 of "LGEBR" and jumble the rest to make the 5 letter word
ways of doing this = C(5,2)*5!/2!

REMOVE A1 and A2 and jumble
=5!

remove one A and one of "LGEBR"
=C(2,1)C(5,1)*5!

total = C(5,2)*5!/2!+C(2,1)C(5,1)*5!+5!=1920

now for the 2nd question, dont you mean we dont want words starting and ending with A?

if so, number of words that starts and ends with A is just P(5,3), ie just jumbling the letters in between

so to count them out, its just 1920-P(5,3)=1860

 

[MONKEYjulz]

Are you sure book's answer is 1860 and not 1680 ?? Which book is this question from?

No book. Just a question from tuition.

total = C(5,2)*5!/2!+C(2,1)C(5,1)*5!+5!=1920

now for the 2nd question, dont you mean we dont want words starting and ending with A?

if so, number of words that starts and ends with A is just P(5,3), ie just jumbling the letters in between

so to count them out, its just 1920-P(5,3)=1860

According to the answers, isn't it just messed up the 1st question is 1260, then the question without A's first and last, is 1860? it's not suppose to go up.

 

[kieranjackson91]

instead of picking which 5 letters to use, start with picking which 2 letters we are not using. And label the As ie first A is A1 and 2nd is A2

Lets remove 2 of "LGEBR" and jumble the rest to make the 5 letter word
ways of doing this = C(5,2)*5!/2!

REMOVE A1 and A2 and jumble
=5!

remove one A and one of "LGEBR"
=C(2,1)C(5,1)*5!

total = C(5,2)*5!/2!+C(2,1)C(5,1)*5!+5!=1920

wouldn't it be 5C2*5! + 5C1*5!+5!=1920
i dont get why there are two ways to remove an A (you say C(2,1) ), and why do you divide by 2! when you 'remove 2 of "LGEBR" and jumble the rest to make the 5 letter word', i mean you are going to have two A's in this case in the set of 5.

just wondering..

 

[k02033]

wouldn't it be 5C2*5! + 5C1*5!+5!=1920
i dont get why there are two ways to remove an A (you say C(2,1) ), and why do you divide by 2! when you 'remove 2 of "LGEBR" and jumble the rest to make the 5 letter word', i mean you are going to have two A's in this case in the set of 5.

just wondering..

there are 2 ways of removing A, because i labeled the As to be A1 and A2, (ie i pretend the 2As are actualyl different) and the 2! comes from unlabeling the As to get rid of extra counts like A1A2 and A2A1 are really the same words. This might sound redundant, but it helps me personally, and in my opinion, it works pretty well for harder questions as well. This is the way that i like to tackle these questions. By no means the best. I think yours is simpler and better! So as long as you got the right answer, you shouldnt worry too much about how i got it.