In how many ways can you arrange the integers {1,2,3,3,4,4,5,6} if...?

EDIT: I ammended my earlier solution, since it contained an error.

The total possible number of arrangements = 8! / (2 * 2!)

This is because you have 2 repeat numbers.

To get the arrangements with the 3s together, consider both of them to be one digit, say 33

Now there are 7 digits, so total arrangements with 3s together

= 7! / 2! because of the repeat 4s.

Similarly with 4s together, 7! / 2! ways.

BUT we have overcounted the cases with both 3s together and 4s together.

The number of ways that can happen, is 6! Once again we consider them as unit digits.

So your answer should be

8!/(2*2!) - (2*7!/2! - 6!) = 5760

NOTE:

you can verify this method on a smaller scale.

With the 4 digits 1,2,2,3

you can count 12 possible arrangements, 6 of which involve the 2s together.

12 = 4! / 2

6 = 3!

Same method works.

FrstGrade's answer is a sturdy heuristic, if a,b,c are completely unconstrained. (different than the plain sanity constraint c ? 0. and because c=0 or a million are trivial strategies, we'd besides say c ? 2 and c nonsquare, different than for {4,9,sixteen...}) in spite of the undeniable fact that it helps extremely if we've *some* bounds on a,b,c; particularly if (say) a can basically be beneficial integer, or a minimum of is bounded with the aid of some variety a_min ? a ? a_max. Then n = a+b?c b?c = (n-a) ?c = (n-a)/b using a_min ? a ? a_max (n-a_max)/b ? ?c ? (n-a_min)/b ((n-a_max)/b)² ? c ? ((n-a_min)/b)² Rounding the LHS and RHS... ceil[ ((n-a_max)/b)² ] ? c ? floor[ ((n-a_min)/b)² ] this components some bounds c_min ? c ? c_max or equivalently you ought to get b_min ? b ? b_max Now you have bounds on a,b,c to your heuristic finding out, i.e. 2 of here: a_min ? a ? a_max b_min ? b ? b_max 2 ? c_min ? c ? c_max (Bounds on c are much less clever than on a or b simply by fact that there will be greater applicants for ?c for given variety of a or b)

Look at it step-by-step.

For the first position, there are six possibilities: 1,2,3,4,5, or 6.

For the next position, one of those possibilities is gone so there are 5. Then four, three, two, one. So far we have 720 possibilities. But we have another 3 and another 4.

There are five places where we can insert one of the remaining digits -- at the beginning, or after any non-equal digit; of course, we can't put it BEFORE any matching digit either. 5*720= 1500 possibilities now.

Then there are six positions where we can now insert the last digit. (see previous paragraph for why.) 1500*6 = 9000 possible permutations and combinations.

OK total number of Combinations is

8!/(2!2!)=10080

Now, the number of arrangements with 2 3s next to each other is

7!=5040

Similarily, 2 4s next to each other is also the same number:

5040

However, we have counted the numbers with 3s and 4s next to each other twice, so we add on

6!=720

The answer is 720

there is 5760 possible combinations. 1x2 x3x4x5x6x8=5,760. If you're not sure how this is done by this example you need to ask the right question

Good Luck

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