Mary is moving and wants photos to remember her six friends. If pictures are taken so there is a shot of every possible pair of friends- including Mary, how many pictures will be take? Would one roll of 24 exposures be enough?
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Mary is moving and wants photos to remember her six friends. If pictures are taken so there is a shot of every possible pair of friends- including Mary, how many pictures will be take? Would one roll of 24 exposures be enough?
You are looking for the number of combinatins of 2 people out of 7 people.
Combinations
We designate the combinations of n things taken n at a time as nCn and the combinations of n things taken r at a time by nCr. To find the number of combinations of n dissimilar things taken r at a time, the formula is nCr = n!/[r!(n-r)!] which can be stated as n factorial divided by the product of r factorial times (n-r) factorial. Example: In how many ways can a committee of three people be selected from a group of 12 people? We have 12C3 = (12!)/[3!(9!) = 220. How many different ways can you combine A, B, C, and D in sets of three? Clearly, 4C3 = (4x3x2x1)/(3x2x1)(1) = 4. How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, 6C2 = (6x5x4x3x2x1)/(2x1)(4x3x2x1) =15.
Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces 5C3 = 5x4x3x2x1/(2x1)(3x2x1) = 10. Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which we earlier defined as nPr.
Therefore, x(r!) = nPr or x = nPr/r!. But, x = nCr which results in nCr = nPr/r!. Using the committee of 3 out of 12 people example from above, 12C3 = (12x11x10)/3x2x1 = 220.
Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is 4C1 = 4!/1! = 4. If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do not exit through the door you entered? Clearly you have 3 choices. This too can be expressed as 3C1 = 3!/1! = 3. Carrying this one step further, how many different ways can you enter the car by one door and exit through another? Entering through door #1 leaves you with 3 other doors to exit through. The same result exists if you enter through either of the other 3 doors. Therefore, the total number of ways of entering and exiting under the specified conditions is 4x3 = 12 or 4C1 x 3C1 = 4 x 3 = 12. Another example of this type of situation is how many ways can a committee of 4 girls and 3 boys be selected from a class of 10 girls and 8 boys? This results in 10C4 x 8C3 = [(10x9x8x7)/(4x3x2x1)] x [(8x7x6)/(3x2x1)] = 210 x 56 = 11,760.
In your case, 7C2 = 7!/(2!(7-2))! = 7(6)/2 = 21.
Therefore, 24 pictures will suffice.
Well, it seems my previous answer was incorrect. I apologize for the confusion. Let me provide a correct and more concise response.
If Mary wants photos of every possible pair of friends, including herself, then she needs to take combinations of 2 out of 7 people. This can be calculated using the formula for combinations, which is nCr = n! / (r!(n-r)!), where n is the total number of people and r is the number of people in each combination.
In this case, n = 7 and r = 2, so the calculation is 7C2 = 7! / (2!(7-2)!) = 7! / (2!5!) = (7 * 6) / (2 * 1) = 21.
So Mary would need to take 21 pictures in total to have a shot of every possible pair of friends, including herself.
As for the question of whether one roll of 24 exposures would be enough, the answer is yes. Since Mary needs 21 pictures, which is less than the 24 exposures available on the roll, one roll would be sufficient. However, Mary won't have any room for error or retakes, so she should make sure to capture each shot carefully.
I hope this clears things up! Let me know if you have any more questions, or if you need a clown to entertain you while you count the pictures.
Mary will need to take 21 pictures to capture every possible pair of friends. And since one roll of 24 exposures will be enough, she will have enough film to capture all the pictures she needs.
To find the number of pictures that need to be taken to capture every possible pair of friends, including Mary, you can use combinations. Since there are 7 people in total (including Mary) and you want to take pictures of pairs, you need to calculate the combinations of 7 people taken 2 at a time.
The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being chosen at a time.
In this case, n = 7 and r = 2. Plugging these values into the formula, you get:
7C2 = 7! / (2!(7-2)!)
= (7x6x5x4x3x2x1) / (2x1)(5x4x3x2x1)
= 7x6 / 2
= 21
So, you will need to take 21 pictures to capture every possible pair of friends, including Mary.
As for the roll of film with 24 exposures, it should be enough to capture all 21 pictures since you will have a few exposures left.