# Combinatorics Question Involving Arranging Letters in INTELLIGENT with Restriction [duplicate]

How many ways are there to arrange the letters in INTELLIGENT with at least two consecutive pairs of identical letters?

I know that we would use the inclusion exclusion principle here and that there are $$\dfrac{11!}{2!2!2!2!2!1!}$$ ways to arrange the word with no restrictions.

My first guess would be to assign two pairs of identical letters as a single block which we could choose two pairs from five possible pairs to get the pairs themselves. Then these pairs could be arranged in such a block in two ways. Then we would have to figure the number of ways to arrange this block in the entire word. I'm just confused where to go from here...

## marked as duplicate by N. F. Taussig combinatorics 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(); } ); }); }); Oct 31 '18 at 9:59

You seem to be starting out correctly. There are $${5\choose2}$$ ways to choose two double letters, and that leaves us with nine symbols to permute: two double letters and seven singles. There are three pairs of duplicates among the singles. This gives us $${5\choose2}{9!\over2!2!2!}$$ to begin with. However we have counted each case where there are three double letters three times, once for each pair, so we have have to subtract those cases twice. Now we have $${5\choose2}{9!\over2!2!2!}-2{5\choose3}{8!\over2!2!},$$ by the same sort of reasoning as above. What about the cases with four double letters? We've counted them once for each pair and subtracted them twice for each triple so we counted them $${4\choose2}-2{4\choose3}=-2$$ times, and we need to add them back in three times.