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Sun 22 May, 2005 06:16 am
Hi there, just got a few questions that i would like some answers to, any help would be appreciated a lot!
Question 1
(i) How many different 8 bit strings can be made from the word kununurra, assuming that each of the nine characters is used at most once?
Question 2
(a) A starting netball team is to be formed from a group of 11 players.
(i) If 2 forwards, 3 centres, and 2 defenders are to be chosen, in how many ways can the team be chosen?
(ii) If there is a captain and vice-captain to be chosen, along with 5 other players and one reserve, in how many ways can this be done?
(b) Use the principle of Inclusion-Exclusion to determine how many integers there are between 100 and 500 inclusive, which are divisible by 3, 5 or 11.
Question 3
There are 12 people in a group. Seven of these are women, and five of these are men. How many different committees of 7 people can be formed if the committee contains:
(i) exactly 4 women
(ii) at least 4 women
(iii) a female president and secretary, a male treasurer, and at most two other males.
Question 4
Cards are drawn from a deck of cards. Determine the probability that:
(i) the first three cards drawn are hearts
(ii) the first three cards drawn are a 2,3 4 or 6
(iii) the sum of the first five cards drawn is less than 16 (an ace counts as 1)
It's these kind of questions that make me feel partially like a failure.I wish I had the insight to mathematically figure these out.
These are in the subject of combinatorial analysis, and use binomial coefficients.
(1)(i)
I assume 8-bit string means 8-character string. I also assume you can't tell one 'n' from another and so on. Therefore, you can omit a 'k', 'u', 'n', 'r', or 'a'. The number of ways to arrange N items with repeated elements is N!/(R1!*R2!*...*Rm!) where R1, R2, ..., Rm are the number of occurrences of each item. R1+R2+...+Rm = N.
omit k: There are 8!/(3!*2!*2!*1!) arrangements
omit u: 8!/(1!*2!*2!*2!*1!)
omit n: 8!/(1!*3!*1!*2!*1!)
omit r: 8!/(1!*3!*2!*1!*1!)
omit a: 8!/(1!*3!*2!*2!)
Now add them all up.
(2)(b)
I don't know if this is using the I-E principle, but the answer will be:
the number of integers divisible by 3
plus the number of integers divisible by 5
plus the number of integers divisible by 11
minus the number of integers divisible by 3*5
minus the number of integers divisible by 3*11
minus the number of integers divisible by 5*11
plus the number of integers divisible by 3*5*11
(2)(a)(i)
There are C(11,7) ways to choose seven players from the group of eleven. Using the same reasoning as in (1)(i), there are 7!/(2!*3!*2!) ways to assign positions FFCCCDD to them.
So, the answer is C(11,7) * 7!/(2!*3!*2!)
which is equivalent to P(11,7)/(2!*3!*2!)
(2)(a)(ii)
The same method applies here. Now there are eight players to choose and positions CVOOOOOR to assign to them.
(3)(i)
How many ways can you select the four women?
How may ways can you select men to fill the remaining positions?
(3)(ii)
It gets a bit trickier. I think this is a simple as selecting four women, then filling the remaining positions from the remaining women and men.
(3)(iii)
P(7,2) ways to select the female pres. and sec.
C(4,1) ways to select the male treasurer
That leaves four positions to be filled with at most two other males. Therefore, there are at least two other females.
How many ways can you fill the two female positions from the remaining females?
Now, how many ways can you fill the remaining two positions from the remaining males and females?
(4)(1)
The probability of three hearts is
(the number of ways three hearts can be drawn) / (the number of ways three cards can be drawn from the complete deck)
There are 13 hearts. How many ways can you draw three?
(4)(ii)
Same reasoning as above.
(the number of ways three cards can be drawn from the set of 12) / (the number of ways three cards can be draw from the complete deck)
133 integers divisible by 3--100mod3=2 (500-100)mod3=1 so # of integers is int((500-100)/3)+0=133
81 integers divisible by 5--100mod5=0, (500-100)mod5=0 so # of integers is int((500-100)/5)+1=81
36 integers divisible by 11 --100mod11=1, (500-100)mod11=4 so # of integers is int((500-100)/11)+0=36
Rap
markr wrote:(1)(i)
I assume 8-bit string means 8-character string. I also assume you can't tell one 'n' from another and so on. Therefore, you can omit a 'k', 'u', 'n', 'r', or 'a'. The number of ways to arrange N items with repeated elements is N!/(R1!*R2!*...*Rm!) where R1, R2, ..., Rm are the number of occurrences of each item. R1+R2+...+Rm = N.
omit k: There are 8!/(3!*2!*2!*1!) arrangements
omit u: 8!/(1!*2!*2!*2!*1!)
omit n: 8!/(1!*3!*1!*2!*1!)
omit r: 8!/(1!*3!*2!*1!*1!)
omit a: 8!/(1!*3!*2!*2!)
Now add them all up.
Yes, all your assumptions are correct. I added them up and got 15120. The thing that always has me questioning whether i have the right answer is the fact that 9!/(3!*2!*2*) also equals 15120, the question to that answer being "How many nine bit strings can be made from kununurra". So i came to the conclusion that to get the number of nine bit strings the first 8 bits have to be unique in order to get all 9 bits that are unique. Is this correct?
I guess it makes sense that they are the same. Arrange the nine letters, then pick the first eight for your string.
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