# Permutations and combinations

• Dec 2nd 2009, 04:58 PM
arze
Permutations and combinations
I have three questions, the first two are similar.
1)Four boxes each containing a large number of identical balls, those in one box are red, those in the second box are blue, those in the third box are yellow, and those in the remaining box are green. In how many ways can a set of five balls be chosen if:
a) there is no restriction
b) at least one ball is red?

For a) i think the answer should be $4^5=1024$ but the answer is 56. For b) i think it should be $4^4=256$ but the answer is 35.

2)In how many ways can four tins of fruit be chosen from a supermarket offering ten varieties if at least two of the tins are of the same variety?

There are at least two cans that are the same, so we have two same, three same, and all same. I did $10\times 9\times 8=720$ for two same, $10\times 9=90$ for three same, and 10 for all same. Total would be 820, but the answer is supposed to be 505.

3)A certain test consists of seven questions, to each of which a candidate must give one of three possible answers. According to the answer that he chooses, the candidate must score 1,2, or 3 marks for each of the seven questions. In how many ways can a candidate score exactly 18 marks in the test?

I don't know how to begin this question.

Thanks for any help!
• Dec 2nd 2009, 06:54 PM
aman_cc
Quote:

Originally Posted by arze
I have three questions, the first two are similar.
1)Four boxes each containing a large number of identical balls, those in one box are red, those in the second box are blue, those in the third box are yellow, and those in the remaining box are green. In how many ways can a set of five balls be chosen if:
a) there is no restriction
b) at least one ball is red?

For a) i think the answer should be $4^5=1024$ but the answer is 56. For b) i think it should be $4^4=256$ but the answer is 35.

2)In how many ways can four tins of fruit be chosen from a supermarket offering ten varieties if at least two of the tins are of the same variety?

There are at least two cans that are the same, so we have two same, three same, and all same. I did $10\times 9\times 8=720$ for two same, $10\times 9=90$ for three same, and 10 for all same. Total would be 820, but the answer is supposed to be 505.

3)A certain test consists of seven questions, to each of which a candidate must give one of three possible answers. According to the answer that he chooses, the candidate must score 1,2, or 3 marks for each of the seven questions. In how many ways can a candidate score exactly 18 marks in the test?

I don't know how to begin this question.

Thanks for any help!

I'm gonna give a quick reply. Will request you to please check if it matches with the ans. If not plz post back

1 a)
8!/(5!3!)
ans is = num of solutions to the eq x1+x2+x3+x4=5, with each of xi >= 0
can you see why?
b) use above to work this out - it is not tough

2) Try this
X = ways to choose tins without any restriction
Y = ways to chose without any repetition

X,Y should be easy to find
so ans is X-Y

3) it is the coefficient of x^18 in the expansion of (x+x^2+x^3)^7

So whats that? Can you see why should this be the ans? Infact Q1 above is done in a similar way

I would suggest you please break down all the problem in some std ones - then understand how to solve the std ones. It helps that way
• Dec 2nd 2009, 11:47 PM
arze
is there any other way to do the first and last question? i haven't learnt that method yet.
Thanks
• Dec 2nd 2009, 11:51 PM
arze
Quote:

Originally Posted by aman_cc
2) Try this
X = ways to choose tins without any restriction
Y = ways to chose without any repetition

X,Y should be easy to find
so ans is X-Y

the number of ways without any restriction would be $10^4$? and the number of ways without any repetition would be $^{10}C_4=210$?
then X-Y would be 10000-210=9790? but this is not the answer.
• Dec 3rd 2009, 01:56 AM
emakarov
For 1a), see the description of combinations with repetitions in Wikipedia. This gives the result suggested above. I find example 2 there easier to understand.
• Dec 5th 2009, 08:57 PM
aman_cc
Quote:

Originally Posted by arze
I have three questions, the first two are similar.
1)Four boxes each containing a large number of identical balls, those in one box are red, those in the second box are blue, those in the third box are yellow, and those in the remaining box are green. In how many ways can a set of five balls be chosen if:
a) there is no restriction
b) at least one ball is red?

For a) i think the answer should be $4^5=1024$ but the answer is 56. For b) i think it should be $4^4=256$ but the answer is 35.

2)In how many ways can four tins of fruit be chosen from a supermarket offering ten varieties if at least two of the tins are of the same variety?

There are at least two cans that are the same, so we have two same, three same, and all same. I did $10\times 9\times 8=720$ for two same, $10\times 9=90$ for three same, and 10 for all same. Total would be 820, but the answer is supposed to be 505.

3)A certain test consists of seven questions, to each of which a candidate must give one of three possible answers. According to the answer that he chooses, the candidate must score 1,2, or 3 marks for each of the seven questions. In how many ways can a candidate score exactly 18 marks in the test?

I don't know how to begin this question.

Thanks for any help!

You are confusing. When you select - the order doesn't matter. So for e.g.

the number of ways without any restriction would be http://www.mathhelpforum.com/math-he...4c72196a-1.gif? and the number of ways without any repetition would be http://www.mathhelpforum.com/math-he...1bd74188-1.gif?
then X-Y would be 10000-210=9790? but this is not the answer.

is worng