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How many permutations exist of the letters a b c d taken four at a time?
Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
How many license plates are possible consisting of 4 letters followed by 3 digit if the first letter must be a C?
1x26x26x26x10x10x10=17576000
How many distinguishable permutations of letters are possible in the word critics?
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.
How many permutations of 3 different digits are there from the ten digits 0 to 9 inclusive?
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.
How many cuboids can you make with the volume of 24Cm cubed?
Infinitely many. Select any number A such that 0 < A ≤ cuberoot(24) = 2.8845 approx. Then let S = 24/A Next, let B be a number such that A ≤ B ≤ sqrt(S) and let C = 24/(A*B) Then it can be shown that A ≤ B ≤ C and a cuboid with sides of length A, B and C will ave a volume of A*B*C = 24 cm3. There are infinitely many possible values for A so that there are infinitely many possible cuboids.
How many six-letter sequences are possible that use the letters c p c c p k?
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.
How many permutations exist of the letters a b c d taken two at a time?
12
How many permutations exist of the letters a b c d taken four at a time?
Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
What is the notation for a permutations?
The Greek letter pi. pi(abcd) represents permutations of the letters in the set {a,b,c,d}.
How do you swap 4 variables?
tmp= a; a= d; d= b; b= c; c= tmp; Other permutations are also possible.
How many permutations are in the word computer?
7. C-O-M-P-U-T-E-R
How many arrangements can be made with the letters sociological?
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.
What letters does not represt one possible bases in the DNA structre a b c g?
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).
How many differrent ways can you arrange the letters of the word revolver?
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.
How many license plates are possible consisting of 4 letters followed by 3 digit if the first letter must be a C?
1x26x26x26x10x10x10=17576000
How many distinguishable permutations of letters are possible in the word critics?
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.
How many distinguishable permutation in Cincinnati?
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.
