Solution to PERMUTATIONS (mod4) 1. There are 6 people to be arranged in a line for a … - Sikademy
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Archangel Macsika

PERMUTATIONS (mod4) 1. There are 6 people to be arranged in a line for a concert. How many arrangements are possible? 2. How many strings of length 5 can be formed using the letters QUALITY if a. Repetitions are not allowed? b. Repetitions are allowed? c. Starts with letter L and repetition is not allowed? 3. A group of 25 people are going to run a race. The top three runners earn gold, silver, and bronze medals. How many arrangements are possible? 4. In how many different ways can the letters of the word "CHANGE" be arranged in such a way that the vowels always come together? 5. Find the number of permutations of the word INFORMATION.

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Here's the Solution to this Question

(1) Answer: 6!= 720 ways

Explanation:

There are 6 ways to choose the first person, then 5 to choose the next, etc.

Hence the number of possible choices is: 6\times 5\times4\times3\times 2\times 1=6!=720

(2) QUALITY

(a) Since no letters of word QUALITY are repeated

There are 7 ways to choose the first alphabet of string, then 6 to choose the next, etc.

So, total number of strings of length 5 = 7\times 6\times 5\times4\times3=2520


(b) If repetitions are allowed

Then, There are 7 ways to choose the first alphabet of string, and 7 to choose the next, etc.

So, total number of strings of length 5 = 7\times 7\times 7\times7\times 7=16807


(c) Start with letter L and repetition is not allowed

Then the string has one alphabet fixed and remaining 4 can be arranged

So, total number of strings = 6\times 5\times 4\times 3=360


(3) Total number of arrangements = ^{25}P_3=13800\ \ arrange ments\ \ possible


(4) CHANGE

Vowels in words are : AE

Then \boxed{AE}CHNG can be arranged as : 2\times 5!=240\ \ ways

SO, there are 240 ways in which the word CHANGE can be arranged such that vowels are always together.


(5) INFORMATION

Number of letters = 11

Repetitions: 'I'=2 times

'N'= 2 times

'0'= 2 times

Hence, Total number of permutations of word = \dfrac{11!}{2!\times2!\times 2!}=4989600


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Question ID: mtid-5-stid-8-sqid-3048-qpid-1747