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In how many ways can the letters of the word GARGANTUNG be
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09 May 2019, 02:07
09 May 2019, 02:07
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56% (01:17) correct
44% (01:12) wrong based on 50 sessions
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In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s appear together?
(A) \(\frac{8!}{3!*2!*2!}\)
(B) \(\frac{8!}{2!*2!}\)
(C) \(\frac{8!*3!}{2!*2!}\)
(D) \(\frac{8!}{2!*3!}\)
(E) \(\frac{10!}{3!*2!*2!}\)
(A) \(\frac{8!}{3!*2!*2!}\)
(B) \(\frac{8!}{2!*2!}\)
(C) \(\frac{8!*3!}{2!*2!}\)
(D) \(\frac{8!}{2!*3!}\)
(E) \(\frac{10!}{3!*2!*2!}\)
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Re: In how many ways can the letters of the word GARGANTUNG be
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09 May 2019, 09:08
09 May 2019, 09:08
5
Carcass wrote:
In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s appear together?
(A) \(\frac{8!}{3!*2!*2!}\)
(B) \(\frac{8!}{2!*2!}\)
(C) \(\frac{8!*3!}{2!*2!}\)
(D) \(\frac{8!}{2!*3!}\)
(E) \(\frac{10!}{3!*2!*2!}\)
(A) \(\frac{8!}{3!*2!*2!}\)
(B) \(\frac{8!}{2!*2!}\)
(C) \(\frac{8!*3!}{2!*2!}\)
(D) \(\frac{8!}{2!*3!}\)
(E) \(\frac{10!}{3!*2!*2!}\)
----ASIDE--------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------------------------
First, "GLUE" the 3 G's together to form ONE super letter (which we'll call X)
So, we must arrange the letters in the word: ARANTUNX
There are 8 letters in total
There are 2 identical A's
There are 2 identical N's
So, the total number of possible arrangements = 8!/[(2!)(2!)]
Answer: B
Cheers,
Brent
General Discussion
Re: In how many ways can the letters of the word GARGANTUNG be
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26 Nov 2019, 23:20
26 Nov 2019, 23:20
GreenlightTestPrep wrote:
Carcass wrote:
In how many ways can the letters of the word GARGANTUNG be rearranged such that all the G’s appear together?
(A) \(\frac{8!}{3!*2!*2!}\)
(B) \(\frac{8!}{2!*2!}\)
(C) \(\frac{8!*3!}{2!*2!}\)
(D) \(\frac{8!}{2!*3!}\)
(E) \(\frac{10!}{3!*2!*2!}\)
(A) \(\frac{8!}{3!*2!*2!}\)
(B) \(\frac{8!}{2!*2!}\)
(C) \(\frac{8!*3!}{2!*2!}\)
(D) \(\frac{8!}{2!*3!}\)
(E) \(\frac{10!}{3!*2!*2!}\)
----ASIDE--------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------------------------
First, "GLUE" the 3 G's together to form ONE super letter (which we'll call X)
So, we must arrange the letters in the word: ARANTUNX
There are 8 letters in total
There are 2 identical A's
There are 2 identical N's
So, the total number of possible arrangements = 8!/[(2!)(2!)]
Answer: B
Cheers,
Brent
What if the letter would be ACCLAIM ?
