Help in Discrete Math

[rebelnole93]

Can anyone help me study for my Discrete math final. I have some problems and would be greatly thankful for any step by step help / explanations on how to solve them.

For #1 a) I got 26 falling 5 and for b)26^5

#2) Determine the number of non-negative integer solutions to the equation

• X₁ + X₂ + X₃ + x₄ = 17

#3) Use inclusion/exclusion counting to find the number of different ways that 5 students could exchange papers for grading so that each student grades exactly one paper, and no student grades his or her own paper

 

[Steven G]

Can anyone help me study for my Discrete math final. I have some problems and would be greatly thankful for any step by step help / explanations on how to solve them.

For #1 a) I got 26 falling 5 and for b)26^5
View attachment 5281View attachment 5282View attachment 5283

Hi, Someone on the forum will certainly help you but you need to do two things first. Can you show us your attempts in doing these problems and can you post the problems so that they can be read?

 

[rebelnole93]

#2

for number 2, I got 20 choose 17, which equals 1140.

 

[pka]

Can anyone help me study for my Discrete math final. I have some problems and would be greatly thankful for any step by step help / explanations on how to solve them.
For #1 a) I got 26 falling 5 and for b)26^5
#2) Determine the number of non-negative integer solutions to the equation

• X₁ + X₂ + X₃ + x₄ = 17

#3) Use inclusion/exclusion counting to find the number of different ways that 5 students could exchange papers for grading so that each student grades exactly one paper, and no student grades his or her own paper

for number 2, I got 20 choose 17, which equals 1140.

You failed to post the question in #1. So NO COMMENT

You are correct about #2.

#3 is a derangement. Study that webpage. You can see the answer here.

 

[sohith]

#2

#2 is a large one

CONSIDER x1,x2,x3,x4 to be 4 non-negative integers ranging from 0-17

x1+x2+x3+x4=17
case 1:-let x1=17
then the rest of the 3 elements are 0
USING PERMUTATIONS AND COMBINATIONS ,the above soln. is possible in 4 ways
(i.e. 17+0+0+0=17 & 0+17+0+0=17 & 0+0+17+0=17 & 0+0+0+17=17)
case 2:-let x1=16
then 1 element is '1' and 2 elements are '0'
USING P&C ,the above soln. is possible in 24 ways
case 3:-let x1=15
then (a)1 element is '2' and the other 2 elements are '0'
(b)1 element is '0' and the other 2 elements are '1'
now in (a) it is possible in 24 ways and in (b) it is possible in 24 ways
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case x:-the above cases are possible only upto x1=9 because from x1=8 values start repeating

NOW LET US CONSIDER ALL POSSIBLE WAYS

4 + 24 + 24^2 + 24^3 +.......+24^8

THE FORM SEEMS LIKE "G.P. SERIES"

BUT 1 IS MISSING ,so

LET "S=(4=3+1)3+1+ 24 + 24^2 + 24^3 +.......+24^8

SUM=S-3=1+ 24 + 24^2 + 24^3 +.......+24^8

S-3=1[(24^8)-1]/[24-1]

S-3=4785883225

S=4785883228

pls let me know if it's right or wrong

 

[pka]

#2 is a large one
CONSIDER x1,x2,x3,x4 to be 4 non-negative integers ranging from 0-17
x1+x2+x3+x4=17
case 1:-let x1=17
then the rest of the 3 elements are 0
USING PERMUTATIONS AND COMBINATIONS ,the above soln. is possible in 4 ways
(i.e. 17+0+0+0=17 & 0+17+0+0=17 & 0+0+17+0=17 & 0+0+0+17=17)
case 2:-let x1=16
then 1 element is '1' and 2 elements are '0'
USING P&C ,the above soln. is possible in 24 ways
case 3:-let x1=15
then (a)1 element is '2' and the other 2 elements are '0'
(b)1 element is '0' and the other 2 elements are '1'
now in (a) it is possible in 24 ways and in (b) it is possible in 24 ways
.case x:-the above cases are possible only upto x1=9 because from x1=8 values start repeating

NOW LET US CONSIDER ALL POSSIBLE WAYS
4 + 24 + 24^2 + 24^3 +.......+24^8
THE FORM SEEMS LIKE "G.P. SERIES"
BUT 1 IS MISSING ,so
LET "S=(4=3+1)3+1+ 24 + 24^2 + 24^3 +.......+24^8
SUM=S-3=1+ 24 + 24^2 + 24^3 +.......+24^8
S-3=1[(24^8)-1]/[24-1]
S-3=4785883225
S=4785883228 s let me know if it's right or wrong

Do you have any ides what that is all about? I think not.
Have a look at this.

This is such an important idea, what a shame for someone to mess with it.
The number of ways to place \(\displaystyle \bf{N}\) identical objects into \(\displaystyle \bf{K}\) distinct cells is \(\displaystyle \dbinom{N+K-1}{N}\).

Historically that is proven by the Stars&Bars method,

 

[sohith]

Do you have any ides what that is all about? I think not.
Have a look at this.

This is such an important idea, what a shame for someone to mess with it.
The number of ways to place \(\displaystyle \bf{N}\) identical objects into \(\displaystyle \bf{K}\) distinct cells is \(\displaystyle \dbinom{N+K-1}{N}\).

Historically that is proven by the Stars&Bars method,

PLS. can u explain it briefly................

ALSO explain the sum

 

[Steven G]

PLS. can u explain it briefly................

ALSO explain the sum

The link that was supplied actually has the proof. Did you look at it? If you did not understand the proof can you please explain where you get lost?