What in the sane heck did you guys do to get perms and combs into your heads? (1 Viewer)

bleakarcher

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Its so frustrating I was trying to get this question:
In how many ways can 4 consonauts and 3 vowels be arranged in a row if (b) so that the first and the last places are occupied by consonauts?
and apparently the answer is 3!*4!*5C3. How?!!!!!!!!!!!!!!!!
 

Shadowdude

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Do it logically:

1. Place a consonant in the first place... 4 ways to pick the consonant
2. Place a consonant in the last place... 3 ways
3. Place the other 5 letters in the other 5 places... 5! ways

5!*4*3 = 3!*4!*C(5,3)
 

bleakarcher

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In these probabiltiy questions I get confused wif when to use factorials, perms, combs. Is it possible do them them all with use of factorials only?
 
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SpiralFlex

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When a human gets confused with Mathematics. Always go back to the beginning and re-read!
 

SpiralFlex

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Once you find the missing link and concept BOOM! Things start to connect.
 

Shadowdude

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In these probabiltiy questions I get confused wif when to use factorials, perms, combs. Is it possible do them them all with use of the Ps and Cs?
Do it logically. I did my solution logically and it worked. I can't be bothered to figuring out how they got 3!*4!*C(5,3) - but the point is 4*3*5! is the same thing.

Other students will go through their working and have 3!*4!*C(5,3) as a final answer.

But I'll go through my answer in full, so you can kinda see where I'm coming from:

1. Place a consonant in the first place... 4 ways to pick the consonant
2. Place a consonant in the last place... 3 ways

At this point we have 5 letters, 2 consonants and 3 vowels to place in 5 spots. But we don't care about whether a letter is a consonant or a vowel anymore. So we proceed as follows:

3. Pick the first consonant and place it in one of the 5 spots available... 5 ways to place it
4. Pick the second consonant and place it in one of the 4 spots available... 4 ways
5. Pick the first vowel and place it in one of the 3 spots available... 3 ways
6. Pick the second vowel and place it in one of the 2 spots available... 2 ways
7. Place the last vowel into the last spot available... 1 way

you could also say:

3. For the first available spot, pick one letter... 5 letters to pick from
4. For the next spot, pick a letter... 4 ways

(and so on)

Answer: 4*3*5*4*3*2*1 = 4*3*(5!)

So you do maths and these questions until you're comfortable. When I started out, I did my working in full like above. But then I realised and got confident with when I was supposed to use factorials, Ps and Cs, so I started using them in my answers as a short cut.
 
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bleakarcher

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Question from mathsonline: The letters EEEBBBB are arranged in a row. How many different arrangements are there? How many different arrangements are there of just 6 of the letters are possible?
 

math man

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For the first part: We have 3 E's and 4 B's in total, i.e 7 letters, so they can be rearranged 7! ways, however since there are 3 E's and 4B's we divide by (3!)(4!)
to cancel the repetition. so total ways = (7!)/(3!4!)=5040/(6x24)=35
For the second part we want to pick only 6 letters, now we must think of all possible cases, luckily there are not many...
Our only options are if we have 3E's and 3B's or 2E's and 4B's, anything other than this then we cant have 6 letters. So we now work out number
of ways for each:

Case1 (3 of each): total ways = (6!)/(3!3!)=(720)/(6x6)=20

Case2 (2 E's 4 B's): total ways = (6!)/(2!4!)=(720)/(2x24)=15

Since we can have case 1 OR case 2 we just simply add the two and get 35.

Now for all these soughts of questions i usually like to lay it out by drawing tile places depending on how many letters...i.e _ _ _ _ _ _
and writing out all possibilites or the ones i dont want and subracting it from total
 

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