<p>A B C D E F G </p>
<p>If the 7 cards above are placed in a row so that cards A and G are at either end, and if B, C and D must be kept together, how many different arrangements are possible?</p>
<p>2<em>3</em>2<em>1</em>1=12
just view BCD as a whole</p>
<p>There are seven slots, but there are limitations as to what can go in each slot.</p>
<p>The first card can either be an A or a G, so there are two choices. Similarily, the last card can only be what is left over from the pair, so there is only one choice.</p>
<p>After that, we have a couple of possibilities on the middle five cards.</p>
<p>Starting with the second slot, there are five options for what that card can be (since BCD does not have to stay in that order.</p>
<p>IF the second card is a B, C, or D, then the third card must be one of that triplet, and the fourth card must be one as well. (We will deal with the other option later.)</p>
<p>Then the fifth card can either be E or F, and the sixth card will be the last.</p>
<p>That’s a multiplication of 2<em>3</em>2<em>1</em>2<em>1</em>1, or 24 arrangements.</p>
<p>Now, what if the second card isn’t a B, C, or D? But the third is one of those, and then the fourth and fifth are similarily a triplet card? The sixth card would still be a single card.</p>
<p>So 2<em>2</em>3<em>2</em>1<em>1</em>1, or another 24.</p>
<p>And then you need to consider if the B, C, D triplet doesn’t come until the fourth, fifth, and sixth cards (2<em>2</em>1<em>3</em>2<em>1</em>1).</p>
<p>I’m getting 24*3, which is 72, mainly because we do not have a limit that BCD stay together *in that order<a href=“so%20CDB%20is%20just%20as%20likely”>/i</a>, etc.</p>
<p>Agree with purple acorn.</p>
<p>6 ways to arrange BCD </p>
<p>2 ways to pick first and last card: A-G or G-A.</p>
<p>6 ways to arrange EF: EF or FE before BCD, or the same after BCD, or one on each side of BCD, with E before BCD or F before.</p>
<p>6 x 2 x 6 = 72</p>
<p>i do agree with #3!! Sorry for the mistakes i’ve made in #2…
and i wanna change my solutions…
2<em>3</em>2<em>1</em>1=12
B,C,D have 6 arrangements
12<em>(3</em>2*1)=72</p>
<p>Thanks guys, yes, the correct answer is 72</p>