Pegs are to be placed in the four holes shown, one in each hole. The pegs come in different colours and pegs of the same colour are identical. Calculate how many different arrangements of coloured pegs in the four holes can be made using
(i) 6 pegs, all of different colours, [1]
(ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg. [1]
Beryl has 12 pegs consisting of 2 red, 2 blue, 2 green, 2 orange, 2 yellow and 2 black pegs. Calculate how many different arrangements of coloured pegs in the 4 holes Beryl can make using
(iii) 4 different colours, [1]
(iv) 3 different colours, [3]
(v) any of her 12 pegs.
Please explain
your answers.
(i) 6 pegs, all of different colours: 6! (6 factorial) = 720 different arrangements. This is because there are 6 pegs and each peg can be placed in any of the 4 holes, giving 6x6x6x6 = 6! possible arrangements.
(ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg: 4! (4 factorial) = 24 different arrangements. This is because there are 4 pegs and each peg can be placed in any of the 4 holes, giving 4x4x4x4 = 4! possible arrangements.
(iii) 4 different colours: 4! (4 factorial) = 24 different arrangements. This is because there are 4 different colours and each colour can be placed in any of the 4 holes, giving 4x4x4x4 = 4! possible arrangements.
(iv) 3 different colours: 3! (3 factorial) = 6 different arrangements. This is because there are 3 different colours and each colour can be placed in any of the 4 holes, giving 3x3x3x3 = 3! possible arrangements.
(v) any of her 12 pegs: 12! (12 factorial) = 479,001,600 different arrangements. This is because there are 12 pegs and each peg can be placed in any of the 4 holes, giving 12x12x12x12 = 12! possible arrangements.
Oh, pegs in holes! That's like a game of hide and seek for colorful little sticks! Let's calculate the number of arrangements:
(i) For the first case, we have 6 pegs of different colors. Since we need to put one peg in each hole, the number of arrangements would be 6! (pronounced "6 factorial," which means multiplying all the numbers from 6 down to 1: 6 × 5 × 4 × 3 × 2 × 1), and that gives us 720 unique arrangements.
(ii) In the second case, we have specific pegs: 2 blue, 1 orange, and 1 yellow. To calculate the number of arrangements, we need to consider that two of the pegs are identical (blue), while the others are unique. Using the formula for counting arrangements of objects with repetition, we get 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 12 arrangements.
(iii) Now for Beryl! She has 12 pegs, consisting of 2 of each color. If we choose 4 different colors to put in the holes, we have 6 options since Beryl has 6 different colors available. Then, using the formula for combinations (nCr), we have 6C4 = (6 × 5 × 4 × 3) / (4 × 3 × 2 × 1) = 15 different arrangements.
(iv) If Beryl only uses 3 different colors, we have to consider different scenarios. We can have either 3 pegs of the same color and 1 of a different color or 2 pegs of one color, 1 of another color, and 1 of another color. Let's calculate both cases:
- 3 pegs of the same color: Beryl has 6 options for the chosen color. Only one color can be chosen, so we have 6 different arrangements.
- 2 pegs of one color, 1 of another color, and 1 of another color: Beryl has 6 options for the first color, 5 options for the second color (as one color has already been chosen), and 4 options for the third color. However, we need to divide by 2 to account for the identical pegs of the first color. Therefore, we get (6 × 5 × 4) / 2 = 60 different arrangements.
(v) Lastly, for any of Beryl's 12 pegs, we have to consider that she can choose any pegs for each hole. This means she has 12 options for the first peg, 12 options for the second peg, 12 options for the third peg, and 12 options for the fourth peg. Using the multiplication principle, we get 12 × 12 × 12 × 12 = 20,736 different arrangements.
I hope that helps with the peg puzzle! If not, let me know, and I'll try to tickle your funny bone with another explanation.
(i) For the first question, since there are 6 pegs of different colors and 4 holes to place them in, we can calculate the number of different arrangements using the formula for permutations.
The number of different arrangements can be calculated as:
6P4 = 6! / (6-4)! = 6! / 2! = (6 x 5 x 4 x 3) / (2 x 1) = 6 x 5 x 4 = 120
So there are 120 different arrangements possible.
(ii) In this question, we have 2 blue pegs, 1 orange peg, and 1 yellow peg. We need to calculate the number of different arrangements of these pegs in the 4 holes. Again, we can use the formula for permutations.
The number of different arrangements can be calculated as:
4P2 = 4! / (4-2)! = 4! / 2! = (4 x 3) / (2 x 1) = 12 / 2 = 6
So there are 6 different arrangements possible.
(iii) For this question, Beryl has 12 pegs consisting of 4 different colors. Since we need to choose 4 pegs to place in the 4 holes, we can calculate the number of different arrangements using the formula for combinations.
The number of different arrangements can be calculated as:
12C4 = 12! / (4! x (12-4)!) = 12! / (4! x 8!) = (12 x 11 x 10 x 9) / (4 x 3 x 2 x 1) = 495
So there are 495 different arrangements possible.
(iv) In this question, Beryl has 12 pegs consisting of 3 different colors. We need to calculate the number of different arrangements of these pegs in the 4 holes. We can use the same formula for combinations as in the previous question.
The number of different arrangements can be calculated as:
12C3 = 12! / (3! x (12-3)!) = 12! / (3! x 9!) = (12 x 11 x 10) / (3 x 2 x 1) = 220
So there are 220 different arrangements possible.
(v) In this question, Beryl has 12 pegs of any color. Since we can choose any of her 12 pegs to place in each hole, the number of different arrangements is simply 12 x 12 x 12 x 12 = 12^4 = 20736. So there are 20736 different arrangements possible.
To calculate how many different arrangements of coloured pegs can be made in the four holes, we'll use a concept called permutations. A permutation is an arrangement of objects in a specific order.
(i) In the first case, we have 6 pegs, all of different colors. To find the number of possible arrangements, we need to find the number of permutations of 6 objects taken 4 at a time.
This can be calculated using the formula: nPr = n! / (n - r)!, where n is the total number of objects and r is the number of objects taken at a time.
For this case, the number of permutations will be 6P4 = 6! / (6 - 4)! = 6! / 2! = 6 × 5 × 4 × 3 = 360.
So, there are 360 different arrangements possible.
(ii) In the second case, we have 4 pegs - 2 blue, 1 orange, and 1 yellow. To find the number of possible arrangements, we again use the permutation formula.
The number of permutations will be 4P4 = 4! / (4 - 4)! = 4! / 0! = 4 × 3 × 2 × 1 = 24.
So, there are 24 different arrangements possible.
(iii) In the third case, Beryl has 12 pegs consisting of 2 red, 2 blue, 2 green, 2 orange, 2 yellow, and 2 black pegs. She wants to choose 4 different colors for the 4 holes.
To calculate this, we use combinations since we are selecting colors without regard to order.
The number of combinations will be 6C4 = 6! / (4! × (6 - 4)!) = 6! / (4! × 2!) = 6 × 5 / (2 × 1) = 15.
So, there are 15 different arrangements possible.
(iv) In the fourth case, Beryl has 12 pegs and wants to choose 3 different colors for the 4 holes.
We can approach this in two ways:
- Option 1: We choose 3 colors and leave one hole empty.
The number of combinations will be 6C3 = 6! / (3! × (6 - 3)!) = 6! / (3! × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 20.
- Option 2: We choose 2 colors and duplicate one of them in two holes.
The number of combinations will be 6C2 × 2 = 6! / (2! × (6 - 2)!) × 2 = 6! / (2! × 4!) × 2 = (6 × 5) / (2 × 1) × 2 = 30.
Hence, in total, there are 20 + 30 = 50 different arrangements possible.
(v) In the last case, Beryl can choose any of her 12 pegs for each of the 4 holes. This means for each hole, she has 12 choices.
So, the number of different arrangements possible is 12^4 = 20736.