Permutation problem - How many permutations are there such that no two numbers are immediately adjecent? [duplicate]

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Consider the set of numbers 1,1,2,2,3,3,4,4.

How many permutations are there such that no two identical numbers are immediately adjacent?

marked as duplicate by Gerry Myerson, Grigory M, Najib Idrissi, Namaste homework StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 17 '14 at 11:10

• We just had exactly this problem yesterday. I'll see if I can find it. Here it is --- math.stackexchange.com/questions/868930/… – Gerry Myerson Jul 17 '14 at 9:46
• I think this is slightly different, if we consider numbers like 12, 31 etc. We want to avoid having 12123434, as there are identical numbers immediately adjacent. I we are talking about one-digit numbers however, it is the same. – Martigan Jul 17 '14 at 9:50
• @Martigan, you may be right --- but as OP writes "numbers" and then lists digits, I choose to interpret "numbers" as meaning "digits". – Gerry Myerson Jul 18 '14 at 6:59

There are $\frac{8!}{2!^4}$ permutations in total. $8$ symbols, in 4 pairs of 2 identical.
There are $\frac{4!}{1!3!}\frac{7!}{2!^3}$ ways at least 1 pair are coadjacent. (Choose the pair, then treat it as one symbol).
There are $\frac{4!}{2!2!}\frac{6!}{2!^2}$ ways at least 2 pair are coadjacent. (Choose the pairs, then treat each as one symbol).
There are $\frac{4!}{3!1!}\frac{5!}{2!}$ ways at least 2 pair are coadjacent. (Choose the pairs, then treat each as one symbol).
There are $4!$ ways all pairs are coadjacent.