3145 and 1543 are the same combination of numbers, but are different permutations. That is, the order of the numbers is different.
When you are holding a hand of cards, only the combination matters.
We designate the permutations of n things taken n at a time as nPn and the permutations of n things taken r at a time as nPr where P stands for permutations, n stands for the number of things involved, and r is less than n. To find the number of permutations of n dissimilar things taken n at a time, the formula is nPn = n! which is n factorial which means n(n-1)(n-2)(n-3).......3x2x1. Example: How many ways can you arrange the letters A & B. Clearly 2 which is 2 x 1 = 2. How many ways can you arrange the letters A, B & C in sets of three? Clearly 3P3 = 3 x 2 x 1 = 6. How many ways can you arrange A, B, C & D in sets of four? Clearly 4P4 = 4 x 3 x 2 x 1 = 24.
To find the number of permutations of n dissimilar things taken r at a time, the formula is nPr = n(n-1)(n-2)(n-3)..........(n-r+1). Example: How many 3-place numbers can be formed from the digits 1, 2, 3, 4, 5, and 6, with no repeating digit? Then we have 6P3 = 6x5x(6-3+1) = 6x5x4 = 120. How many 3-letter arrangements can be made from the entire 26 letter alphabet with no repeating letters? We now have 26P3 = 26x25x(26-3+1) = 26x25x24 = 15,600. Lastly, four persons enter a car in which there are six seats. In how many ways can they seat themselves? 6P4 = 6 x 5 x 4 x (6-4+1) = 6x5x4x3 = 360.
Another permutation scenario is one where you wish to find the permutations of n things, taken all at a time, when p things are of one kind, q things of another kind, r things of a third kind, and the rest are all different. Without getting into the derivation, nPn(p,q,r,s) = n!/(pxqxr). For example, how many different permutations are possible from the letters of the word committee taken all together? There are 9 letters of which 2 are m, 2 are t, 2 are e, and 1 c, 1 o, and 1 i. Therefore, the number of possible permutations of these 9 letters is 9P9(2,2,2,1,1,1) = 9!/(2x2x2x1x1x1) = 362,880/8 = 45,360.
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.