permutations and combinations. Am I right?

Question

1. Frisbees come in 5 models, 8 colors, and 3 sizes. How many Frisbees must the local dealer have on hand in order to have one of each kind available?

I got 120.

2. A craftperson has six different kinds of seashells. HOw many different bracelets can be constructed if only four shells are to be used in any one bracelet?

I got 360

If a person is dealth two cards from a 52-card deck. How many different hands are possible if order is not important?

I got 2652.


Con someone help with these?

4. How many odd numbers of three digits each can be formed from the digits 2,4,6,7 if repitition of digits is permitted?

5. Given eight points in a plane, no three of which are colinear. How many lines do the points determine?

Answer 1

You got the first 3 correct. For 4. I think the answer is 3*2=6. because the 7 has to come last. For problem 5. the answer is 7+6+5+4+3+2+1=28. To understand this, try placing these 8 points on the vertices's of a regular octagon.

Answer 2

Number 4 is not 6 since repetition of numbers is allowed.
Any one of the 4 numbers can be the 1st digit. Any one of the 4 nunbers can be the 2nd digit. Only the 7 can be the last digit. Thus the total number of odd numbers that can be formed is 4*4*1 = 16

Your answers to the first 3 appear correct.

D= n(n-3)/2 is formula to get number of diagonals of a polygon of n sides. Using n=8, we get D = 8(8-3)/2 = 20 which should be the answer to question 5.

Answer 3

1) right

2) it say how many different bracelet. a,b,c,d and the same as a,c,b,d. What you found is permutation.

6! / ((6-4)! * 4!) = 15

3) again what you found is permutation. 1 and 2 is the same as 2 and 1. 2652 / 2 = 1326 ways.

4) since you want a odd number, 7 has to be the last number.
there are 3 numbers left. 3x2x1 = 6 ways

you still want a odd number. again, 7 has to be the last number. 1x3x3x3 = 27 ways

5) it's combination. 8! / ((8-2)! * 2!) = 28 lines

I'm not very sure of my answers because i still need helps on combination and permutation. But its good that i try to answers and learn it.

Answer 4

1. 120 ok
2. 360 ok .. I take it that the different shells are used in different ways so that the order of choosing is important.
3. the order of the cards in the hand is not important so 52*51/2 = 1326
4. choose from 4 for hundreds, choose from 4 for tens, can only choose 7 for units as it must be odd ..= 4*4 = 16

5. a line is defined by any pair of points and order not important... so 8 * 7 / 2 = 28

Answer 5

enable her elect the ribbons one via one. For the 1st one, she has 6 concepts. For the 2d, there are 5 concepts left. So there are 30 diverse procedures of choosing an ordered pair of ribbons. although, Cathy would not care with reference to the order. Taking first a blue and then a white ribbon is comparable to taking a white ribbon first and then a blue one. this implies we counted double the style of opportunities. So the honestly huge style is 5*6 / 2 = 15. this is, in effect, a mix: 2C6. The reasoning in the back of the formulation for mixtures is as follows: you are able to positioned all six ribbons in 6! diverse orderings. Of the six appeared after ribbons you may then choose the 1st 2. As you do no longer care with reference to the order of those 2, you are able to divide the huge style via the style of procedures of ordering those 2: 2!. a similar is going for the 4 ribbons Cathy did no longer choose: those would properly be in 4! diverse orderings. So the whole style of obtainable mixtures is 6! / (2! 4!)

Answer 6

1. 5*8*3=120
2. npr(6,4) if order is important
3. ncr(52,2)=1326 order not important