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Nội dung chính Show- Video TranscriptHow many permutations are there of the letters?How many different arrangements are there of the letters in the word correct?How many different letter arrangements can be made from the letters in the word statistics?How many different arrangements are there for the 11 letters in the word combination?
VerifiedHint: To solve this question we need to have the knowledge of permutation and combination. The first step is to calculate the number of times each letter has occurred to form the word. Then as per the condition write all the cases required to form a word which question demands for.
Complete step by step answer:
The question asks us the number of ways in which a five letter word could be made with exactly three different letters from the word CURRICULUM. To solve this the
first step will be to divide the problem in cases for making the calculation easier.
On analysing we see that in letter CURRICULUM, “C” occurs twice, “U” occurs thrice , “R” occurs twice , “I” occurs once , “L” occurs once , “M” occurs once. The cases that will be formed are:
Case1: For exactly three different letters all the “U” will be used with any two letters which are different like (“C”,”I”,“I”, “L”, “M”). Number of ways to select these letters are:
$Rightarrow ^5C_2=dfrac5!2!left(
5-2 right)!$
$Rightarrow ^5C_2=dfrac5!2!left( 3 right)!$
$Rightarrow ^5C_2=dfrac5times 4times 3!2!left( 3 right)!$
$Rightarrow ^5C_2=10$
The number of arrangement is =$dfrac5!3!times 10=200$
Case 2: Considering 2”C” and 2”R” to form a word of five letter. Number of ways to select these letters are:
$Rightarrow ^4C_1=dfrac4!1!left( 4-1 right)!$
$Rightarrow ^4C_1=dfrac4!1!left( 3
right)!$
$Rightarrow ^4C_1=4$
The number of arrangement is =$dfrac5!2!2!times 4=120$
Case 3: Considering 2”C” and 2”U” to form a word of five letter. Number of ways to select these letters are:
$Rightarrow ^4C_1=dfrac4!1!left( 4-1 right)!$
$Rightarrow ^4C_1=dfrac4!1!left( 3 right)!$
$Rightarrow ^4C_1=4$
The number of arrangement is =$dfrac5!2!2!times 4=120$
Case 4: Considering 2”U” and 2”R” to form
a word of five letter. Number of ways to select these letters are:
$Rightarrow ^4C_1=dfrac4!1!left( 4-1 right)!$
$Rightarrow ^4C_1=dfrac4!1!left( 3 right)!$
$Rightarrow ^4C_1=4$
The number of arrangement is =$dfrac5!2!2!times 4=120$
Now we will be adding the arrangements of all the four cases. On doing this we get:
$Rightarrow 200+120+120+120$
$Rightarrow 560$ ways
$therefore $ Number of ways in which $5$letter
words can be formed using the letter from the word “W” if each $5$ lettered word has exactly $3$different letters, is Option $B)560$.
So, the correct answer is “Option B”.
Note: To solve the question we need to remember the formula for the combination of a number. So the term $^nC_r$ is written as $dfracn!r!left( n-r right)!$ . The term $n!$ means that the number is being multiplied to the natural numbers below $n$. This means $n!=ntimes left( n-1 right)times left( n-2 right)........3times 2times 1$.
Explanation:For problems, like these, we need to consider the number of total letters and the number of repeated letters.
There are 9 letters in this word, so if all the letters were different there would be #9!# ways of arranging them.
However, we have #3# R's, #2# A's and #2# E's.
So, this expression becomes #(9!)/(3! xx 2! xx 2!) = (362,280)/(6 xx 2 xx 2) = (362,280)/24 = 15,120#.
Hopefully this helps!
Video Transcript
In this problem late NBD total number of later which is then repeating later. See two times you three times and uh, good times now. Total distinct permutation. Therefore, total interesting. Total distinct permutation is equal to factorial 10 by Bacterial to multiplication. Factorial three multiplication. Factorial Which is equal to 1512. Do do do do as the final answer for this problem. I hope you understand the solution of this problem.
CURRICULUM = 10 letters
there is 2 C's, 2 R's and 3 U's
therefore
10!/(2!*2!*3!) = 151,200 permutations
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How many permutations are there of the letters?
To calculate the amount of permutations of a word, this is as simple as evaluating n! , where n is the amount of letters. A 6-letter word has 6! =6⋅5⋅4⋅3⋅2⋅1=720 different permutations.How many different arrangements are there of the letters in the word correct?
Hence, the number of different arrangements in the word CORRECT is 1260.How many different letter arrangements can be made from the letters in the word statistics?
Therefore, the letters in the word "STATISTICS" can be arranged in 50400 distinct ways.How many different arrangements are there for the 11 letters in the word combination?
Answer: COMBINATION can be arranged in 4,989,600 different ways if it is eleven letters and only use each letter once. Tải thêm tài liệu liên quan đến nội dung bài viết How many different permutations can be made from the letters of the word CURRICULUM