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Solved Examples

Example-1
In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
  1. 6!/2!
  2. 3! * 3!
  3. 4!/2!
  4. (4!*3!)/2!
  5. (3!*3!)/2!
Solution
ABACUS is a 6 letter word with 3 of the letters being vowels.
 
If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together. These 4 elements can be rearranged in 4! Ways.
 
The 3 vowels can rearrange amongst themselves in (3!/2!) ways as "a" appears twice.
 
Hence, the total number of rearrangements in which the vowels appear together are (4!*3!)/2!
 
  
Example-2

How many different four letter words can be formed (the words need not be meaningful) using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

  1. 59
  2. 11! / (2!*2!*2!)
  3. 56
  4. 23
  5. 11! / (3!*2!*2!*2!)
Solution
The first letter is E and the last one is R..Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
 
The second and third positions can either have two different letters or have both the letters to be the same.
 
Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.
 
Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.
 
Total number of posssibilities = 56 + 3 = 59
 
  
Example-3
What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re-arranged?
  1. ΒΌ
  2. 1/6
  3. 1/3
  4. 1/24
  5. 1/12
Solution
The total number of ways in which the word Math can be re-arranged = 4! = 4*3*2*1 = 24 ways.
 
Now, if the positions in which the consonants appear do not change, the first, third and the fourth positions are reserved for consonants and the vowel A remains at the second position.
The consonants M, T and H can be re-arranged in the first, third and fourth positions in 3! = 6 ways without the positions in which the positions in which the consonants appear changing.
Therefore, the required probability = 3!/4! = 6/24 = 1/4
 
 
Example-4

There are 6 boxes numbered 1, 2, ....6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

  1. 5
  2. 21
  3. 33
  4. 60
  5. 6
Solution
If only one of the boxes has a green ball, it can be any of the 6 boxes. So, this can be achieved in 6 ways. If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
 
Similarly, if 3 of the boxes have green balls, there will be 4 options.
 
If 4 boxes have green balls, there will be 3 options.
 
If 5 boxes have green balls, then there will be 2 options.
 
If all 6 boxes have green balls, then there will be just 1 options.
 
Total number of options = 6 + 5 + 4 + 3 + 2 + 1 = 21.
  
 
Example-5
A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?
  1. 1
  2. 1/256
  3. 81/256
  4. 175/256
  5. 144/256
Solution
The man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes. So, the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes.
 
The probability that he will not hit the target in one shot = 1 – 1/4 = 3/4
 
Therefore, the probability that he will not hit the target in all the four shots = ΒΎ*3/4*3/4*3/4 = 81/256
 
Hence, the probability that he will hit the target at least in one of the four shots = 1 – 81/256 = 175/256.
 
 
Example-6

In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?

  1. 5C3
  2. 5P3
  3. 53
  4. 35
  5. 25
Solution
The first letter can be posted in any of the 3 post boxes. Therefore, it has 3 choices.
 
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes. Therefore, the total number of ways the 5 letters can be posted in 3 boxes is 3*3*3*3*3
 
= 35
 
  
Example-7
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?
  1. 210
  2. 29
  3. 3 * 28
  4. 3 * 29
  5. None of these
Solution
When a coin is tossed once, there are two outcomes. It can turn up a head or a tail. When 10 coins are tossed simultaneously, the total number of outcomes = 210
 
Out of these, if the third coin has to turn up a head, then the number of possibilities for the third coin is only 1 as the outcome is fixed as head. Therefore, the remaining 9 coins can turn up either a head or a tail = 29
 
 
Example-8

In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat?

  1. 7!
  2. 7C7
  3. 77
  4. 49
  5. None of these
Solution
There are seven positions to be filled. The first position can be filled using any of the 7 letters contained in PROBLEM.
 
The second position can be filled by the remaining 6 letters as the letters should not repeat.
 
The third position can be filled by the remaining 5 letters only and so on.
 
Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! Ways
 




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