The product principle is that if there are m different ways of operating and for each of these, there are n different ways of performing a second independent operation. There are mn different ways of performing the two operations in succession.
Here we got m and n. We can also have n with the respective subscripts.
A more sophisticated phrase to this storyline is "Combinatorial analysis."
A fun prob...: The wireframe below forms the box’s outline. An ant crawls along the wire from P to Q. In how many different ways of the shortest length lead from P to Q?
From Q, there are 3 possible first-leg ways to A, B, or C. There are 2 second-leg ways to P. Totally, 6 paths.
Upon a board, there are 10 distinct signs in total. How many ways are needed to reflect 4 yellow signs? 10×9×8×7=5040
If there are 6 tickets that go into the hands of 3 people, how many ways to do so? 3×3×3×3×3×3=9×9×9=729
You got 7, 11, and 13 pebbles of different colors respectively. In how many ways could you choose the first? 7 possible col1 to consider and so the first pick can be carried out in 7 ways. 11 possible col2 to consider and so the second pick can be carried out in 11 ways. 13 possible col3 to consider and so the third pick can be carried out in 13 ways. Apply the multiplication (or product) principle
Consider the first 4 letters out of the 26. How many ways is this possible if they are in clockwise alphabetical order, they are to be in alphabetical order, and they are to be in random order?
Firstly, there are 4 choices for A, but if A is fixed, 1 for B, 1 for C, and 1 for D, so 4×1×1×1=4 ways. Secondly, there are four choices for A, but if A is fixed, its 2 choices for B, and B is fixed, its 1 for C and 1 for D. Multiply them as usual. Thirdly, there are 4 choices for W but if A is fixed, 3 choices for B and B is fixed, 2 for C, 1 for D. Multiply as usual.
If there are 3 digits with one non-zero digit in it and 2 different letters, how many different vehicle registration number plates would be accepted? Considering the fact that they're different from one another, 26 different ways for the first letter, (26-1) different ways for the second letter, and for the digits part, 9 ways for that first non-zero digit, 10 ways for the next immediate digit, and the same for the final one(I mean, any of the 10 numbers can be chosen unlike the non-zero scenario). Multiply as usual.
Consider the digits 9,8,7,6,5 and 4.
Figure out the number of five-digit numbers that can be formulated from those 6 digits. And hello, isn't this in the nCr format? There are 6 digits to decide from, in order to extract some 5-digit numbers from it. Anyway, as many times, as you want alone is a different scenario. That'd be straight-off 6×6 = 36 ways. Whereas, if there is that only once each condition, 5!=5×4×3×2×1=60
By the way,
n! = 1×2×3....×(n-2) (n-1) n
0! = 1
And – multiply the possibilities Or – add the possibilities
Permutation
Chosen all at one time, It's an arrangement of a set of objects in a specific order. And when it's the permutation of n number of objects chosen r at a time, it's the r <= n (in a specific order anyway). This follows that n gotta be greater than r. Observably, this time, the order matters.
P(n,r) is the number of permutations of n number of objects taken (how many) r at a time
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