Math Help! Permutations and Combinations?

Question

Erm.. help me please. :D


1. How many 4-letter permutations can be made from the letters PENCIL?

2. How m any 3-letter permutations can be made from the letters COMPUTER?

3. There is a glass of 24 students. The class is lining up in pairs to see a presentation. How many ways could the students group themselves if the order in which they are standing does not matter?


Explain as much as possible please.. or my teacher will start asking me questions. o_o;;

Answer 1

For permutations, the exact order of the elements are important.
For combinations, the exact element are considered, not the order.
Pencil = 6, that is, it has six different letters or elements.
6^P_4 = 6*5*4*3 = 360
Computer = 8.
8^P_3 = 8*7*6 = 336

Order does not matter in a combination.
24^C_2 = [24*23]/ 1*2 = 276

Answer 2

1. All the letters in PENCIL are different, you have thus 6 choices for the letter 1, 5 - for letter 2, 4 - for letter 3 and 3 - for letter 4. So you have 6*5*4*3 = 360 possible permutations.

2. Same stuff. All the letters are different, total 8 letters, hence 8*7*6 choices = 336 possible permutations.

3. More difficult. You have to divide everyone in pairs. For the first pair you have 24*23 choices, but as it doesn't matter is the pair is George - Mary, or Mary - George, you shoud've divided them by 2, hence 12*23. For the next pair 22*21/2, etc. Means it will be 24!/(2^12). (All the numbers from 1 to 24 multiplied and then divided by 2 12 times). However, this way we will get stuff like
.... Pair AB ... Pair CD ...
and
.... Pair CD ... Pair AB ...
counted twice - and so on for all pairs. As it doesn't really matter, what is the order in which the pairs stand, you have to divide their number by the number of all possible orders for a given set of pairs, that is 12! (there are 12 pairs total, so 12*11*...*2*1 possible orders).

The complete answer will be then 24!/ [(12!)*(2^12)] which is... (better use calculator!)
3794809718700.

Hope, it's OK for you and your teacher!

Answer 3

You can consider permutation of letters in a very simple way.
1. Assume you have four places to fill since you have to make a permutations of 4-letters.

a)The first place can be filled in 6 ways i.e any of P,E,N,C,I,L
b) Now, once the first place is filled with a letter there remains only 5 letters. Thus, the second place can be filled in 5 ways.
c) Likewise, third place in 4 ways
d) fourth place in 3 ways

Thus, no. of permutions possible will be 6* 5*4*3 =360

The direct formula for it is nPr , where n is the distinct letter in a word and r is the r-letter permutions required

So, 6P4 = 6!/ (6-4)! = 6!/2! = 3*4*5*6 = 360

2) 8P3 = 8!/(8-3)! = 8!/5! = 6*7*8 = 336


3) When order of standing does not matter, we need to find the number of combinations of 2(pairs) which can be formed by 24 students.
24C2 = 24!/(24-2)! 2! = 24!/ 22!2! = 23*24/ 2 = 276

Answer 4

1. If the order in which you arrange the 4 letters matters, then the answer is:

7!/(7-4)! = 7*6*5*4 = 840

If the order doesn't matter, then:

7!/((7-4)!*4!) = 7*5 = 35

2) Same idea: Matters:
8!/(8-3)! = 8*7*6 = 336

Doesn't matter: 8!/((8-3)!3!) = 8*7 = 56

3) Imagine lining them up regularly. There would be 24! different ways of doing it. Now if we divide them into pairs, (1,2)(3,4) etc. we've counted a combination twice for each pair. Since there are 12 pairs, we must divide by 2^12.

So 24!/12!2^12 = a big number

Answer 5

1.
If letters may not be reused,
7*6*5*4 = 840

2.
same here
8*7*6 = 336

3.
Even though they are lining up in pairs, there are 24 positions to be occupied by 24 students.
24! = 620,448,401,733,239,439,360,000