# Simple (ish) discrete maths help

• May 9th 2009, 09:14 AM
rak
Simple (ish) discrete maths help
Hi everyone,

Would really appreciate some help for these questions, could use a pointer in the right direction;

• How many different bridge hands are there which contain 5 clubs, 4 diamonds, 3 hearts and 1 spade?
• How many different bridge hands which contain 5, 4, 3 and 1 card in the four suits in any order?
• The probability of a bridge hand (13 cards) not containing any card of a suit?
• How many ways three distinct numbers can be chosen from {1,2,3,4,5,6,7,8,9,10} so that no two are consecutive?
• How many ways in the range 1 - 1million are perfect squares, or divisible by 4 or 7?

Also if anybody could explain why f(x,y)=x^2+y^2+5 can be represented primitive recursively by SSSSS(A(M(x,x),M(y,y)) that'd be great!

Thanks!
• May 9th 2009, 09:34 AM
TiRune
How many different bridge hands are there which contain 5 clubs, 4 diamonds, 3 hearts and 1 spade?

number of ways to pick 5 out of 13 clubs * number of ways to pick 4 out of 13 diamonds* number of ways to piuck 3 out of 13 hearts* number of ways to pick 1 out of 13 spades.
if you define a bridge hand to be non-reliant on order that is. so you have 4 sub problems. if you have to pick n out of m without order, that is n over m.

How many different bridge hands which contain 5, 4, 3 and 1 card in the four suits in any order?
take the previous example, and do it times the amount of ways to rearrange the 4 suits to the 4 numbers. Should be straightforward.

The probability of a bridge hand (13 cards) not containing any card of a suit?
This is the total amount of bridge-hands, divided by the total amount of bridge hands of only three colors. exactly the same way as the previous two questions, just define how much cards you can pick from and how much are in the bridge hand you want ignoring order.

How many ways three distinct numbers can be chosen from {1,2,3,4,5,6,7,8,9,10} so that no two are consecutive?

You can view this problem as follow. The amount of ways three distinct numbers can be chosen from {1,2,3,4,5,6,7,8,9,10} so that no two are consecutive is equal to: the amount of three distinct numbers MINUS the amount of ways you can pick three distinct numbers and two are consecutive PLUS the amount of ways you can pick three distinct numbers and ALL THREE are consecutive (these get counted twie in the two consecutive + free). The first should be easy, and you can solve the two consecutive and three consecutive by 'glueing' these together. instead of {,}{,}{,}{,}{,}{,}{,}{,}{,}{,} you get {,,,}{,}{,}{,}{,}{,}{,}{,}, {,}{,,,}{,}{,}{,}{,}{,}{,} etc., and the same goes for the two case but with an 'extra' freely to choose number.

How many ways in the range 1 - 1million are perfect squares, or divisible by 4 or 7?

If you mean how many numbers the question is rather simple: just take the root of 1 million, floor it and that's the amount of perfect squares. same goes for the divisions, just divide and floor. If you have to find the amount of numbers that satisfy ANY of these conditions, just add these and substract the numbers that have divisible by 4 or 7 in common (hint: divisible by 4*7), have perfect squares and divisible by 4 in common (hint: any even squared) and perfect squares and divisble by 7 in common (hint: 7 is prime), and PLUS 2*that have all three in common.

what do you mean by SSSSS(A(M(x,x),M(y,y))?

I think these questions belong in the high school forum and not the university forum though ;)
If you need solutions for these questions instead of these overly directive directions feel free to ask.
• May 9th 2009, 10:15 AM
rak

So the answer for the first part would be: 13C5 * 13C4 * 13C3 * 13C1;

Second part: 4! * (13C5 * 13C4 * 13C3 * 13C1)?

And then: 52C13/39C13

Fourth part: 10*9*9 = 810

Fifth part: I ended up with 35917?

And believe it or not, this is a second year university module... I'm okay with the slightly tricker parts of the course but struggle getting my head round these kind of questions!

Thanks again
• May 9th 2009, 10:27 AM
TiRune
Fourth part: 10*9*9 = 810 Seems off, but correct me if I'm wrong: the amount of ways to get 3 numbers out of 10 is 10C3 = 120. the amount of ways to pick a sequenced pair from the 10 numbers is 9 (the amount of commas if u want), and the amount of ways you can order the 1 free number out of the remaining 8 numbers is 8C1 = 8.
so the awnser should be 10C3 - 9*8C1 = 120 - 48 + 8 = 80?

the 7 comes from the fact that the sequenced pair 2,3 with 4 as a free and the sequenced pair 3,4 with 2 as a free are identical so were substracted twice on accident

I get 357643 as an answer for 5 after a fast calculation, how did you get the number?:
root(1000000) = 1000
1000000/4 = 250000
1000000/7 = 142857
1000000/(4*7) = 35714
1000/2 = 500
1000/7 = 142
1000/14 = 71

1000+250000+142857-35714-500-142+2*71 = 357643

The rest seems ok like that, but I am by no means an expert at chance theory :P
• May 19th 2009, 12:35 PM
XzLoB
I believe answer to fourth part is 56.
10C3 total number of ways to select triple
9*8 number of ways to select triple with at least to consecutive numbers
8 ways to select three consecutive numbers (Same method as by TiRune, just his arithmetic went wrong somewhere)

SSSSS(A(M(x,x),M(y,y)) just means: add 1, add 1, add 1, add 1, add 1 to sum of( x*x and y*y) which is x^2 + y^2 + 5 which is composition of primitive recursive functions S is basis function and A,M could easily be shown to be primitive recursive
• May 19th 2009, 01:12 PM
TiRune
Totally right, 9*8 is not 48, wtf o_O