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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?

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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.

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