Permutations and Combinations help! (2 Viewers)

[luvergal]

I hate this chapter. I'm really bad at interpreting questions so I'm having trouble with a lot of questions...even the "easy" ones.

Could someone please help me with these questions? I'm about to rip my head out =O

1. A combination lock is made with 5 dials and 8 numbers on each. If it takes 7 seconds to try each combination how long will it take to unlock the lock by trial anderror assuming the last answer is correct?

2. What is the probability that in a random arrangement of the letters of the word EIGHT, the vowels will come together at the ends?

3. A committee of 6 is to be chosen from 8 wmen and 6 men so as to contain at least 3 women and 2 men.
In how many ways can it be done if two particular women refust to serve together?

Thanks in advance!~

 

[Sandchairs]

i could be wrong since we havent done this yet but i believe the first question goes like this:

5 dials with 8 numbers on each
8!x5 for posible combinations then
8!x5x7 for total time taken
= 392 hours... some ridiculos ammount of seconds

question 2
treat each element as a seperate entity
ie the vowels EI
the rest GHT
so the vowels can be arranged
EI,IE or 2! ways
GHT can be arranged 3! ways
so i get 2!x3!x2 (cause it could be at either end)= 24 ways

Question 3
your calculator has a nCr button (at least mine does i dunno how to actually work it out mathmaticaly)

u have 8 women and want 3 so u press 8C3 (stands for 8 choose 3)
and 6 men chosing 2 so 6C2

(i really dunno whats goin on here but i think based on a diagram i drew)
8C3+6C2-3! (get the team then subtract combinations where 2 women are together)

i dunno how close i am... i suspect im nowhere near so someone else plz confirm/deny
GL with perms

 

[kony]

1.

there are 8 possible numbers in each dial.

hence, 8^5 combinations.

the number of seconds follows.

 

[pLuvia]

1. If there are 8 numbers for each dial, then there are 8 ways of choosing one number for one dial so 8⁵, hence total amount of time is
8⁵*7=229376

2. 2!*2!*3!/5!=1/5

3. Case wise
- 3W 3M
- 4W 2M

Since 2 women don't want to serve together then split them apart from the rest, either one or none of them want to be on the committee

For case 1 3W 3M
=7C3*6C3
=700

For case 2 4W 2M
= 7C4*6C2
=525
Hence total ways=1225

 

[lolokay]

3. A committee of 6 is to be chosen from 8 wmen and 6 men so as to contain at least 3 women and 2 men.
In how many ways can it be done if two particular women refust to serve together?

Thanks in advance!~

I can think of two ways to do this question.

First way: total cominations - cominations w/both refusing women (women A and B)

Ch: 3W, 3M
8C3*6C3 - 6C1*6C3 = 1000
Ch: 4W, 2M
8C4*6C2 - 6C2*6C2 = 825
Total = 1825

Or, firstly treat A and B as the same woman, AB:
Ch: 3W, 3M
7C3*6C3 = 700
Now, 3/7 of these, or 300, will contain AB. Since there is 2 ways to pick from AB, these combinations will be counted twice. This brings the number of cominations to 1000
Ch: 4W, 2M
7C4*6C2 = 525.
4/7 of these, = 300, will be counted twice, so there are 825 combinations,
The total is therefore 1825