WebJul 3, 2016 · Explanation: There are a total of 10 letters. If they were all distinguishable then the number of distinct arrangements would be 10!. We can make them distinguishable by adding subscripts: If we remove the subscripts from the letter O 's, then it no longer makes any difference what order the O 's are in and we find that 1 2! = 1 2 of our 10 ... WebFeb 10, 2016 · In a word where no letters are repeated, such as FRANCE, the number of distinguishable ways of arranging the letters could be calculated by 5!, which gives 120. However, when letters are repeated, you must use the formula n! (n1!)(n2!)... Explanation: There are 4 s's, 3 a's and a total of 9 letters. 9! (4!)(3!) = 362880 24× 6 = 2520
Counting distinguishable arrangements Notes on …
Web1- How many distinguishable arrangements of the energy quanta are possible for the cold block? (Hint: A single atom can have more than one quantum of energy.) 2- What can you … WebThat amounts to finding all the distinguishable arrangements of the 10-letter "word" "ATHEMATICS" and putting an M in the beginning of each. "ATHEMATICS" is a 10-letter "word" and it contains 2 indistinguishable A's 2 indistinguishable T's Thus, using the rule, we divide 10! by (2!)(2!) Answer: How many of the arrangements in part a have the T ... how many humpback whales
SOLUTION: How many arrangements are there of the word
WebHow many arrangements are there of the word MATHEMATICS? Rule: Start with the factorial of the number of letters in the word. Then, for each indistinguishable letter in the word, … WebFind the number of distinguishable arrangements of the letters of the word TRILLION 10080 A panel containing four on-off switches in a row is to be set. Assuming no restrictions on individual switches, use the fundamental counting principle to find the total number of possible panel settings. 16 WebIf we choose m elements from n in a certain order, it is an arrangement. For example, the arrangement of 2 from 3 is АВ, and ВА is the other arrangement. The number of arrangements of m from n is. Example: For the set of А, В and С, the number of arrangements of 2 from 3 is 3!/1! = 6. Arrangements: АВ, ВА, АС, СА, ВС, СВ howard bears running back