8 people have to be seated in a row. Find the number of ways in which this can be done if person X and person Y must always sit together. Input your answer as sum of last 3 digits of number of ways.
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I plan to create a set with lots of combinatorics problems. Wasn't this 2 easy? :P
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This was easy except for the following part:
as sum of last 3 digits of number of ways.
7!*2=10,080, which wasn't correct. How was the answer supposed to be entered?
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It said "the sum of the last three digits". 0+8+0=8
Image link ?@Agnishom
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Hmm? Which image?
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I Mean please look at the note in ques
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@Krishna Ar – That is really a clever question. I'll update my solution
We can do it like this as well. First, we have to find in how many ways we can arrange 6 people. And that is 6!
Now, we can put the 2 persons X and Y in between 2 people and in front and last in 2 different ways, i.e (X,Y);(Y,X) for every permutation.
Therefore, 6! * 14 = 7! * 2!
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But X won't like it if Y sits in the front and he sits in the back. Would he?
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Yeah, did the same thing
that's simply 7!x2! = 10080 since X and Y will become 1 then there are 7 counts in 2 alternation:
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First, instead of trying to find the permutations of 8 people, think of X and Y to form a single entity.
The number of ways we can arrange A, B, C, D, E, F, {X, Y} is 7!
But then again X and Y can sit as (X,Y) or (Y,X). So, the number of ways we can arrange them all is 7 ! × 2 !
(The image in the problem shows the idea of grouping which I have used. If you have followed my solution, you should understand. Philosophically speaking, X and Y are so close and love each other that they are inseperable)