Help Needed: Questions on Counting Techniques and Probability

[F.B]

I'm really sorry for posting all these questions but i really need help.

1. Show that

C 2(2n-1, n-1) = C (2n, n)

Heres what i did.

2 x (2n-1)!/2n-1-n-1)!(n-1)! = 2n!/n

I don't know how to get the LS to equal the RS.

2. Determine the probability of each of the following.

a) a five card poker hand dealt from a standard deck of 52 playing cards results a full house (3 of a kind and 2 of a kind)
b)Two face cards are drawn in a row (w/o replacement) from a standard deck of 52 playing cards given that the first card is a king.
c) a committee of 6 people randomly chosen from seven males and eight females is either all male or female
d) in a six person sprint, Jessie finishes first, Marnie second and Raul last.

Heres what i did.

a)52!/3! x 2! but i think that's wrong. then i did 5 over that total but they have 6 as a numerator and i can't get that.
b)this one i don't kno how to do.
c) C(7,0) + C (8,0)
Is that right?

3. Use the appropriate counting techniques

a)In how many permutations of the digits 123456789 are the number 1 and 2 beside each other.
b)A school has 480 girls and 520 boys. How many committees of 5 members can be formed if there must be atleast 1 boy on each committe.
c) How many groups consisting of at least 2 people can be chosen from a group of 10 people.

a) P(9,2) again its not right and i don't know what to do

b) C (480,4) x C (520, 1) + C (480, 3) x C (520, 2) + C (480,2) x C (520, 3) + C (480, 1) x C (520, 4) + C (480,0) x (520, 5)
I think I am totally off on this one.

c) 10!/(10-2)!
Is this right?

Again sorry for all these questions but i really need to understand this.

 

[HallsofIvy]

I'm really sorry for posting all these questions but i really need help.
1. Show that
C 2(2n-1, n-1) = C (2n, n)
Heres what i did.
2 x (2n-1)!/2n-1-n-1)!(n-1)! = 2n!/n
I don't know how to get the LS to equal the RS.

?? I assume that C(n,i) is the binomial coefficient but what is
"C 2(2n-1,n-1)? Do you mean 2C(2n-1,n-1)? If so, you still need to be more careful how you write things.
C(2n-1,n-1)= (2n-1)!/(2n-1-(n-1)!(n-1)! which is
(2n-1)!/(2n-1-n+1)!(n-1)!, not what you wrote. And C(2n,n) is
(2n)!/(n!(2n-n)!= (2n)!/(n!n!), again, not what you wrote.
Of course, 2n-1-n+1= n so
C(2n-1,n-1)= (2n-1)!/n!(n-1)!. Since (2n-1)! means (2n-1) times (2n-2) times ... down to 1 and (n-1)! means (n-1) times (n-2) times... down to 1 we can cancel (you should be used to doing that with binomial coefficients) everything except (2n-1)times (2n-2)... down to n.
That is, C(2n-1,n-1)= ((2n-1)(2n-2)... (n))/n!

On the right hand side, C(2n,n)= (2n)!/(n!n!)= ((2n)(2n-1)... (1))/(n!(n(n-1)...1)(
and again, we can cancel: C(2n,n)= ((2n)(2n-1)...(n+1))/n!
Now divide C(2n,n) by C(2n-1,n-1), cancel everything you can and see what happens.

2. Determine the probability of each of the following.
a) a five card poker hand dealt from a standard deck of 52 playing cards results a full house (3 of a kind and 2 of a kind)
52!/3! x 2! but i think that's wrong. then i did 5 over that total but they have 6 as a numerator and i can't get that.

It can't be right because it doesn't even take into account the fact that there are four cards of each kind! Don't just plug numbers into formulas. Think!
How could you draw a full house? One way is this. Take any card from the deck- it doesn't matter which one: there are probability of that is 1.
Now what is the probability that the next card will be the same? There are now 51 cards left in the deck and three of them are the same as the first card. The probability that the second card is the same as the first is 3/51. What is the probability that the third card is the same as the first two? How many cards are there left in the deck? How many are the same as the first two? Divide to find the probability of that!
We now have "three of a kind". To get a full house, the next card must NOT be the same as the first three. There are now 49 cards left in the deck, only one of which is the same as the three in our hand so 48 are NOT the same. The probability that we will get a new card is 48/49.
What is the probability that the last card will be the same as the fourth?
There are now 48 cards left in the deck, three of which are the same as the fourth: 3/48.
You now have five probabilities: 1, 3/51, 2/50, 48/49, 3/49. To find the probability that "this andthat and..." all happen multiply them all together. That gives the probability of being dealt AAABB in that order. To take into account all possible orders, calculate how many ways those 5 cards (those 5 letters AAABB) can be ordered. This is where the binomial coefficient comes in. Multiply that by the product of those five fractions.

b)Two face cards are drawn in a row (w/o replacement) from a standard deck of 52 playing cards given that the first card is a king.
this one i don't kno how to do.

This is the easiest one in the group!
You've already drawn a King- which is a face card. To draw "two face cards in a row" you only need to draw another face card. How many cards are left in the pack? How many of them are face cards? (Face cards, of course, are Jack, Queen, King of any suit.

c) a committee of 6 people randomly chosen from seven males and eight females is either all male or female
C(7,0) + C (8,0)
Is that right?

It obviously can't be right! The problem asked for a probability and probabilities are always between 0 and 1. Also I don't see where you have taken into account the fact that the committee consists of 6 people!
First ignore the question of male or female. There are a total of 15 people to choose from. There are 15 ways to choose the first person, then 14 ways to choose the second, 13 ways to choose the third, 12 ways to choose the fourth, 11 ways to choose the fifth, and 10 ways to choose the sixth. There are a total of (15)(14)(13)(12)(11)(10) (which can also be written 15!/9!= 15!/(15-6)!= P(15,6)) ways to choose the committee without any restrictions. That is in a particular order, so we divide by 6! because we don't what to count order: that gives
15!/(6!(15- 6)!)= C(15,6). That is the number of equally likely outcomes.
Now how many ways could you choose the 6 people from only the 7 men? That is where I think you got your C(7,0) but that's not quite correct. What you want is C(7,6) (which would be the same as C(7,1)) but not C(7,0)).
There are then C(8,6) ways to choose all 6 from the 8 women. That is, there are a total of C(8,6)+ C(7,6) to choose a committee that consists onlyof men or only of women.
Finally, divide that by the number of equally likely ways to choose the committee from all people calcualted above.

d) in a six person sprint, Jessie finishes first, Marnie second and Raul last.
?? There doesn't appear to be a question here! If the question is "how many different ways could the race have ended?", since you are given the positions of Jessie, Marnie, and Raul, the only question is where the other 3 finished. How many possible permutations are there of 3 things?

3. Use the appropriate counting techniques
a)In how many permutations of the digits 123456789 are the number 1 and 2 beside each other.
a) P(9,2) again its not right and i don't know what to do

Think of "12" as being a single digit, call it, say "a". How many permutations are there of the 8 symbols a3456789?
But "1 and 2 beside each other" could mean "21" also. Okay, let "b" represent "21". How many permutations are there of the 8 symbols b3456789?
How many permutations altogether?

b)A school has 480 girls and 520 boys. How many committees of 5 members can be formed if there must be atleast 1 boy on each committe.
b) C (480,4) x C (520, 1) + C (480, 3) x C (520, 2) + C (480,2) x C (520, 3) + C (480, 1) x C (520, 4) + C (480,0) x (520, 5)
I think I am totally off on this one.

Since there must be a boy on the commitee, choose a boy first. There are 520 ways of doing that. You now have left 480+ 519= 999 students. How many committees of 4 members can be chosen from them?
Multiply those together to find the number of committees in which a boy was chosen first. Finally, divide by 5 to account for the fact that that boy could have been chosen at any point.

c) How many groups consisting of at least 2 people can be chosen from a group of 10 people.
c) 10!/(10-2)!
Is this right?

No, once again you confusing the binomial coefficient, the number of ways to choose i things from a group of n, with the number of ways to permute i things from a group of n.
Actually the way I would do this is think: there are 10 ways to choose the first person. That leaves 9 people to choose from so there are 9 ways to choose the second person. There are 10(9)= 10!/8!= 10!/(10-2)!
= P(10,2) ways to choose 2 people in that particular order. Since order is not important here and there are 2! ways to order 2 people we need to divide by 2: the answer is 10!/(2!(10-2)!)= C(10,2).
It would be best to actually calculate that and write the answer as a single number!

OOPS! I just notice that the problem said at least 2 people, not exactly two people! Okay, since we know that C(10,2) is the number of groups of two people chosen from the 10, we also know that there are C(10,3) groups of three people, C(10,4) groups of 4, etc. up to C(10,10) groups of 10 people. That last is equal to 1, of course. The only such group is all 10 people. Add all if those together to get the number of groups of "at least 2" people.
Indeed, I can tell you immediately that the answer 2¹⁰- 1- 10= 1024-11= 1013! Do you know what the sum of all binomial coefficients C(n,i) with n fixed and i going from 0 to n is? Do you see that here we are adding all except C(10,0) and C(10,1)? And those are easy to find.

Again sorry for all these questions but i really need to understand this.

 

[F.B]

thx HallsofIvy.

But i still don't get how to do 2c and 3a