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03 Oct 2019, 01:09
Kudos
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The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Hi team
Can anyone help me to resolve my following doubt please?
Using the counting method solving indirectly:
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60 (Ok I get this)
No of ways two I's can be together 4!=24 (Hold on Shouldn't it be 48?)
(I 2! I) _ _ _
4 * 3 * 2 * 1
= 2! * 4! = 48
I, I (either one of the I's can be selected first)
Thank You,
Dablu
A) 12
B) 18
C) 24
D) 36
E) 48
Hi team
Can anyone help me to resolve my following doubt please?
Using the counting method solving indirectly:
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60 (Ok I get this)
No of ways two I's can be together 4!=24 (Hold on Shouldn't it be 48?)
(I 2! I) _ _ _
4 * 3 * 2 * 1
= 2! * 4! = 48
I, I (either one of the I's can be selected first)
Thank You,
Dablu
13 Oct 2019, 04:13
Kudos
Bookmarks
Hello Dablu,
You have got the first part right. The total number of arrangements that can be made by taking all the letters of the word DIGIT is \(\frac{5!}{2!}\) i.e. 60.
This is a question on arrangement of objects and here, the order of arrangement matters. This means, if you can rearrange the Is and get a distinct arrangement, then you could multiply by 2!.
But, we know that rearranging the Is will not give us a different arrangement. Therefore, the total number of arrangements when the Is are together is equal to 4! (since we take the two Is as one object and there are 3 more distinct objects).
Lastly, this is NOT a selection question, Dablu. This is a permutations question. Questions on formation of words from alphabets are permutations questions. Think about it, you used the permutations concept while calculating the total number of words, right? You cannot suddenly switch your approach to selections and say that I can select the Is in two ways.
By the way, since the Is are identical, the number of ways in which you can select ONE of them is ONE, not 2!.
Hope that helps!
You have got the first part right. The total number of arrangements that can be made by taking all the letters of the word DIGIT is \(\frac{5!}{2!}\) i.e. 60.
This is a question on arrangement of objects and here, the order of arrangement matters. This means, if you can rearrange the Is and get a distinct arrangement, then you could multiply by 2!.
But, we know that rearranging the Is will not give us a different arrangement. Therefore, the total number of arrangements when the Is are together is equal to 4! (since we take the two Is as one object and there are 3 more distinct objects).
Lastly, this is NOT a selection question, Dablu. This is a permutations question. Questions on formation of words from alphabets are permutations questions. Think about it, you used the permutations concept while calculating the total number of words, right? You cannot suddenly switch your approach to selections and say that I can select the Is in two ways.
By the way, since the Is are identical, the number of ways in which you can select ONE of them is ONE, not 2!.
Hope that helps!
13 Oct 2019, 04:42
Kudos
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