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THE NUMBER OF WAYS IN WHICH AN EXAMINER CAN ASSIGN 30 MARKS TO 8 QUESTIONS GIVING NOT LESS THAN 2 MARKS FOR EACH QUESTION IS

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736
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THE NUMBER OF WAYS IN WHICH AN EXAMINER CAN ASSIGN 30 MARKS TO 8 QUESTIONS GIVING NOT LESS THAN 2 MARKS FOR EACH QUESTION IS

Guest Apr 8, 2015

#1
+92196
+8

I DON'T KNOW

but

If you put 2 on each question to start with that is 16 points,

the question becomes,

How many ways can you allocate 14 marks to 8 questions.

14,0,0,0,0,0,0,0                              8 ways

13,1,0,0,0,0,0,0                 8!/6! =  56

12,2                                            56

12,1,1                          8!/(5!2!) =168

11,3                                           56

11,2,1                                   $${\frac{{\mathtt{8}}{!}}{{\mathtt{5}}{!}}} = {\mathtt{336}}$$

11,1,1,1,                                $${\frac{{\mathtt{8}}{!}}{\left({\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{280}}$$

10,4                                           56

10,3,1                                      336

10,2,2                                        $${\frac{{\mathtt{8}}{!}}{\left({\mathtt{5}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{168}}$$

10,2,1,1                                   $${\frac{{\mathtt{8}}{!}}{\left({\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{840}}$$

10,1,1,1,1,

9,5

9,4,1

9,3,2

9,3,1,1

9,2,1,1,1

9,1,1,1,1,1

8,6

8,5,1

8,4,2,

8,4,1,1,

8,3,3

8,3,2,1

8,3,1,1,1,

8,2,2,2

8,2,2,1,1

8,2,1,1,1,1,

8,1,1,1,1,1,1,

7,7

7,6,1

7,5,2

7,5,1,1

7,4,3,

7,4,2,1

7,4,1,1,1,

7,3,3,1

7,3,2,2

7,3,2,1,1,

7,3,1,1,1,1,

7,2,2,1,1,1,

7,2,1,1,1,1,1,

7,1,1,1,1,1,1,1,

6,5,3

6,5,2,1

6,5,1,1,1

6,4,4,

6,4,3,1

6,4,2,2,

6,4,2,1,1,

6,4,1,1,1,1,

6,3,3,2

6,3,3,1,1

6,3,2,2,1

6,3,2,1,1,1,

6,3,1,1,1,1,1,

6,2,2,2,2

6,2,2,2,1,1

6,2,2,1,1,1,1,

6,2,1,1,1,1,1,1

5,5,4

5,5,3,1

5,5,2,2

5,5,2,1,1,

5,5,1,1,1,1,

5,4,4,1

5,4,3,2

5,4,3,1,1

5,4,2,1,1,1

5,4,1,1,1,1,1

5,3,3,3

5,3,3,2,1

5,3,3,1,1,1,

5,3,2,1,1,1,1

5,3,1,1,1,1,1,1,

5,2,2,2,2,1

5,2,2,2,1,1,1,

5,2,2,1,1,1,1,1,

4,4,4,2

4,4,4,1,1,

4,4,3,3

4,4,3,2,1

4,4,3,1,1,1,

4,4,2,1,1,1,1,

4,4,1,1,1,1,1,1,

4,3,3,3,1

4,3,3,2,2,

4,3,3,2,1,1

4,3,3,1,1,1,1,

4,3,2,2,2,1

4,3,2,2,1,1,1,

4,3,2,1,1,1,1,1,

4,2,2,2,2,2

4,2,2,2,2,1,1

4,2,2,2,1,1,1,1,

3,3,3,3,2

3,3,3,3,1,1

3,3,3,2,2,1

3,3,3,2,1,1,1,

3,3,3,1,1,1,1,1,

3,3,2,2,2,2,

3,3,2,2,2,1,1,

3,3,2,2,1,1,1,1,

3,2,2,2,2,2,1

3,2,2,2,2,1,1,1,

2,2,2,2,2,2,2

2,2,2,2,2,2,1,1

TOTAL=

OBVIOUSLY THERE IS A MUCH SIMPLER WAY BUT MAYBE THIS WOULD WORK TOO :)     LOL

Melody  Apr 8, 2015
Sort:

#1
+92196
+8

I DON'T KNOW

but

If you put 2 on each question to start with that is 16 points,

the question becomes,

How many ways can you allocate 14 marks to 8 questions.

14,0,0,0,0,0,0,0                              8 ways

13,1,0,0,0,0,0,0                 8!/6! =  56

12,2                                            56

12,1,1                          8!/(5!2!) =168

11,3                                           56

11,2,1                                   $${\frac{{\mathtt{8}}{!}}{{\mathtt{5}}{!}}} = {\mathtt{336}}$$

11,1,1,1,                                $${\frac{{\mathtt{8}}{!}}{\left({\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{280}}$$

10,4                                           56

10,3,1                                      336

10,2,2                                        $${\frac{{\mathtt{8}}{!}}{\left({\mathtt{5}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{168}}$$

10,2,1,1                                   $${\frac{{\mathtt{8}}{!}}{\left({\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{840}}$$

10,1,1,1,1,

9,5

9,4,1

9,3,2

9,3,1,1

9,2,1,1,1

9,1,1,1,1,1

8,6

8,5,1

8,4,2,

8,4,1,1,

8,3,3

8,3,2,1

8,3,1,1,1,

8,2,2,2

8,2,2,1,1

8,2,1,1,1,1,

8,1,1,1,1,1,1,

7,7

7,6,1

7,5,2

7,5,1,1

7,4,3,

7,4,2,1

7,4,1,1,1,

7,3,3,1

7,3,2,2

7,3,2,1,1,

7,3,1,1,1,1,

7,2,2,1,1,1,

7,2,1,1,1,1,1,

7,1,1,1,1,1,1,1,

6,5,3

6,5,2,1

6,5,1,1,1

6,4,4,

6,4,3,1

6,4,2,2,

6,4,2,1,1,

6,4,1,1,1,1,

6,3,3,2

6,3,3,1,1

6,3,2,2,1

6,3,2,1,1,1,

6,3,1,1,1,1,1,

6,2,2,2,2

6,2,2,2,1,1

6,2,2,1,1,1,1,

6,2,1,1,1,1,1,1

5,5,4

5,5,3,1

5,5,2,2

5,5,2,1,1,

5,5,1,1,1,1,

5,4,4,1

5,4,3,2

5,4,3,1,1

5,4,2,1,1,1

5,4,1,1,1,1,1

5,3,3,3

5,3,3,2,1

5,3,3,1,1,1,

5,3,2,1,1,1,1

5,3,1,1,1,1,1,1,

5,2,2,2,2,1

5,2,2,2,1,1,1,

5,2,2,1,1,1,1,1,

4,4,4,2

4,4,4,1,1,

4,4,3,3

4,4,3,2,1

4,4,3,1,1,1,

4,4,2,1,1,1,1,

4,4,1,1,1,1,1,1,

4,3,3,3,1

4,3,3,2,2,

4,3,3,2,1,1

4,3,3,1,1,1,1,

4,3,2,2,2,1

4,3,2,2,1,1,1,

4,3,2,1,1,1,1,1,

4,2,2,2,2,2

4,2,2,2,2,1,1

4,2,2,2,1,1,1,1,

3,3,3,3,2

3,3,3,3,1,1

3,3,3,2,2,1

3,3,3,2,1,1,1,

3,3,3,1,1,1,1,1,

3,3,2,2,2,2,

3,3,2,2,2,1,1,

3,3,2,2,1,1,1,1,

3,2,2,2,2,2,1

3,2,2,2,2,1,1,1,

2,2,2,2,2,2,2

2,2,2,2,2,2,1,1

TOTAL=

OBVIOUSLY THERE IS A MUCH SIMPLER WAY BUT MAYBE THIS WOULD WORK TOO :)     LOL

Melody  Apr 8, 2015
#3
+92196
+5

I think according to Nauseated, the answer is

http://web2.0calc.com/questions/how-many-ways-are-there-to-distribute-12-unlabeled-b***s-into-9-labeled-boxes

Hey Nauseated, Chris and I are still waiting for you to walk us through the logic behind this formula.  :/

14 unlabelled marks into 8 labelled questions

$$\dbinom{13}{7}=\dfrac{13!}{7!6!}= 1716 \;ways$$

That has a lot more chance of being correct!

Melody  Apr 8, 2015
#4
+92196
+5

I still don't get these.  What have I done wrong ??

If I add up all my possibilities that I have listed and the permutations as well, the number would be way higher!

I have listed all the mark possibilities but not all the permuations of them.

What am I doing wrong ??

Melody  Apr 9, 2015
#5
+5

LET a +b+c+d+e+f+g+h =30 , also a,b,c,d.....>orequal to 2

let p=a-2 , q=b-2 , r=c-2 ........

now , (p+2)+(q+2)+......=30

so , p+q+r+......=14

therefore total number of solutions is (14+8-1)C(8-1) =(21)C(7)

I HOPE YOU UNDERSTOOD

Guest Apr 9, 2015
#6
+92196
0

Thanks anon,

Idk

Does any one else want to weigh in here ?

I know I need to study the answers to the original question.  I still haven't spent as much time considering those answers as I want to.    :/

Melody  Apr 9, 2015

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