Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
1x26x26x26x10x10x10=17576000
The word critics has 7 letters which can be arranged in 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 ways. To see this, imagine placing one of the 7 letters in the first slot. Then place a different letter in the second slot (there are only 6 letters left now), then 5 and so on down to 1. We multiply because we must do each of these steps to create a rearrangement of the word critics. The problem then is that critics contains 2 'c's and 2 'i's, which are indistinguishable. For example, we might count "rtiicsc" several times by switching the places of the 'i's or the 'c's even though we cannot tell the difference in the word. So, we must divide out the repetition, by dividing by 2! = 2 * 1 twice. This corrects the over-counting from the duplicate letters. So the correct result is 7!/(2! * 2!) = 1260 distinguishable permutations.
ABC A = 1,2,3,4,5,6,7,8,9; 9 possibilities B = 0,1,2,3,4,5,6,7,8,9; 10 possibilities C = 0,1,2,3,4,5,6,7,8,9; 10 possibilities So there are 9*10*10 = 900 numbers with 3 digits.
The question can be re-stated as asking for the total number of permutations that can be derived from the following two groups of digits: AAABCD and AABBCD, where A, B, C and D are different. The number of ways of choosing the digit A, to be used three times, out of the ten digits {0, 1, 2, ... 9} is 10. Having done that, the number of ways of selecting 3 from the remaining 9 digits is 9C3 = (9*8*7)/(3*2*1) = 84. Thus there are 10*84 = 840 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 3 of the digits are the same, these permutations are not all distinct. In fact, there are 6!/3! = 120 distinct permutations. That makes a total of 840*120 = 100800 such numbers. Next, the number of ways of choosing the digits A and B, each to be used twice, out of the ten digits {0, 1, 2, ... 9} is 10C2 = (10*9)/(2*1) = 45. Having done that, the number of ways of selecting C and D from the remaining 8 digits is 8C2 = (8*7)/(2*1) = 28. Thus there are 45*28 = 1260 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 2 pairs of these digits are the same, these permutations are not all distinct. In fact, there are 6!/(2!*2!) = 180 distinct permutations. That makes a total of 1260*180 = 226800 such numbers. The grand total is, therefore, 100800 + 226800 = 327600 6-digit numbers made from 4 distinct digits.
I'm going to assume you mean combinations - the unique set of these letters in any order with no sequence repeated. With these letters, there are 60 possible combinations. To see the maths behind this, try typing "permutations of {c,c,c,p,p,k}" into wolfram alpha.
12
Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
The Greek letter pi. pi(abcd) represents permutations of the letters in the set {a,b,c,d}.
tmp= a; a= d; d= b; b= c; c= tmp; Other permutations are also possible.
There are 12 factorial or 479,001,600 permutations of the letters in the word SOCIOLOGICAL. However, since the letter O occurs three times, the letter I occurs twice, the letter L occurs twice, and the letter C occurs twice, you have to divide that by 8 (O), 2 (I), 2 (L), and 2 (C) giving you 7,484,400 distinctpermutations.
The letters a, b, and c do not represent possible bases in DNA structure. The four possible bases in DNA are adenine (A), thymine (T), cytosine (C), and guanine (G).
7. C-O-M-P-U-T-E-R
Revolver has eight letters. Assuming you do not care if you make words, there are 40,320 permutations for arranging those eight letters, if you do not count the repeated e's and v's and r's8! = 40,320 = (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8)We must take this result and divide out the repeated elements. Since there are 2 e's, 2 v's and 2 r's:40,320/(2!2!2!) = 5040.
1x26x26x26x10x10x10=17576000
The word critics has 7 letters which can be arranged in 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 ways. To see this, imagine placing one of the 7 letters in the first slot. Then place a different letter in the second slot (there are only 6 letters left now), then 5 and so on down to 1. We multiply because we must do each of these steps to create a rearrangement of the word critics. The problem then is that critics contains 2 'c's and 2 'i's, which are indistinguishable. For example, we might count "rtiicsc" several times by switching the places of the 'i's or the 'c's even though we cannot tell the difference in the word. So, we must divide out the repetition, by dividing by 2! = 2 * 1 twice. This corrects the over-counting from the duplicate letters. So the correct result is 7!/(2! * 2!) = 1260 distinguishable permutations.
If the first and second C are indistinguishable, then there are 554,400 permutations. If one is upper case and the other is lower case, then there are twice as many.