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In how many ways can three different letters be selected from the letters of the word THREE?    
  • a)
    4
  • b)
    5
  • c)
    10
  • d)
    24
  • e)
    60
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In how many ways can three different letters be selected from the lett...
Step 1: Understand the objective
The objective of the question is to find the number of ways in which three different letters can be SELECTED from a set of 4 letters: T, H, R and E.
(Please note that the word THREE has 5 alphabets, but of these, E occurs twice. So, we get only 4 different alphabets from the word THREE: T, H, R and E)
Step 2: Write the objective equation enlisting all tasks
This is a Selection question. Order doesn’t matter here. So we will solve it by using the nCr formula.
In this case, only one task needs to be completed to accomplish the objective: Select 3 letters out of the given set of 4 different letters.
So, the objective equation becomes:
(Number of ways of selecting 3 letters out of 4 different letters) = 4C3
Step 3: Determine the number of ways of doing each task
By using the formula:
We get:
  Upon simplifying this expression, you get 
                                                     4C3=4
 Step 4: Calculate the final answer
 By putting this value in the objective equation, we get:
(Number of ways of selecting 3 letters out of 4 different letters) = 4
Answer: Option A
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Most Upvoted Answer
In how many ways can three different letters be selected from the lett...
To solve this problem, we need to find the number of combinations of three different letters that can be selected from the letters of the word "THREE".

The word "THREE" has five letters, but we are only interested in selecting three different letters.

Step 1: Identify the letters available for selection
The letters available for selection are T, H, R, E, and E.

Step 2: Determine the number of ways to select three different letters
To determine the number of ways to select three different letters, we can use the concept of combinations.

The formula for combinations is given by:
C(n, r) = n! / (r!(n-r)!)

Where n is the total number of items and r is the number of items to be selected.

In this case, n = 5 (the total number of letters available for selection) and r = 3 (the number of letters to be selected).

Plugging the values into the formula, we get:
C(5, 3) = 5! / (3!(5-3)!)
= 5! / (3! * 2!)
= (5 * 4 * 3!) / (3! * 2 * 1)
= (5 * 4) / (2 * 1)
= 20 / 2
= 10

Therefore, there are 10 different ways to select three different letters from the letters of the word "THREE".

Correct answer: c) 10
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This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts: i. First, we will find out all the seven lettered words from the letters of word CLASSIC ii. Next, we will find out how many of these words will have the two Cs together. The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is: = 7 x 6 x 5 x 4 x ... x 1 = 7! Notice that there are two Cs and two S in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2! x 2! So, the total number of words formed is: 7!/(2! x 2!) We need to find how many of these words will have the two Cs together. To do this, let us treat the two Cs as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S which can be done by dividing 6! by 2!. Total number of 7 lettered words such that the two Cs are always together = 6!/2! The fraction of seven lettered words such that the two Cs are always together is: = (number of words with two Cs together/total number of words) = (6!/2!)/(7!/[2! x 2!]) = (2!/7) = 2/7 Hence the correct answer choice is A.

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In how many ways can three different letters be selected from the letters of the word THREE? a)4b)5c)10d)24e)60Correct answer is option 'A'. Can you explain this answer?
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