There are two pairs of letters in the word preamble which have the same number of letters between them in the word as there are between the same letters in the English alphabet.(1) The A and the second E, with three letters between them:p.r.e.A.m.b.l.E........A.b.c.d.E...(2) The first E and the B, with two letters between them:p.r.E.a.m.B.l.e..a.B.c .d.E.....
example of control number in cebuana llhul ier
The number of 3-letter codes that can be formed without repetition is equal to the number of choices for the first letter (4 letters), multiplied by the number of choices for the second letter (3 letters), multiplied by the number of choices for the third letter (2 letters). Therefore, the total number of 3-letter codes that can be formed is 4 x 3 x 2 = 24.
1 tillion
There are too many letters to count. Each soldier wrote to his/her family, describing their experiences, and ther feelings toward the war. :)
The exact number of letters was never specified.
20
Only the number 8 (VIII)
5! 5 * 4 * 3 * 2 *1 = 120 arrangements you take the number of letters in the words and make it a factorial.
1
3
It depends on the length of each word in letters (characters). The average length of a word in a document is 5 letters - shorter words appear more often. -- The answer above contains 125 characters: 119 letters, 1 number, and 5 punctuation marks, for a total of 27 words and 1 number.
There are two pairs of letters in the word preamble which have the same number of letters between them in the word as there are between the same letters in the English alphabet.(1) The A and the second E, with three letters between them:p.r.e.A.m.b.l.E........A.b.c.d.E...(2) The first E and the B, with two letters between them:p.r.E.a.m.B.l.e..a.B.c .d.E.....
The number of permutations of the letters of the word depends upon the number of letters in the word and the number of repeated letters. Since there are nine letters, if there were no repetitions, the number of ways to rearrange these letters would be 9! or 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1. But don't do the multiplication just yet. To account for the repeated letters, we need to divide by 3! (for the 3 Ns) by 2! (for the 2Es) and by another 2! (for the 2 Ss). This gives a final answer of 15,120 permutations of these letters.
BASKETBALL = 10 letters B=2 A=2 S=1 K=1 E=1 T=1 L=2 Number of permutations = 10! /2!2!2! = 10x9x8x7x6x5x4x3x2 / 2x2x2 =3,628,800 / 8 =453,600
The number of different ways the letters of a word can be arranged, when all the letters are different, is the same as the number of permutations of those letters. In this case, the answer is 5!, or 120.
1) Personal Identification Number 2) Postal Index Number (India)