<p>v, w, x, y, z are placed in a row so that x is not at either end, how many different arrangements are possible?</p>
<p>the answer is 72.</p>
<p>please explain this one. i worked this out a couple of times, but i keep getting 96...maybe you can find my error?</p>
<p>thanks</p>
<p>The different combinations when x can be anywhere is 5!, which is 120.</p>
<p>Since x is limited to only 3 of the 5 spots, just multiply 120 by 3/5.</p>
<p>or just his method. lol</p>
<p>It took me five minutes but i figured this out. I will try to explain it as simple as possible.
So you have five letters (v,w,x ,y and z) and you need to find out how many different arrangements are possible ,if x is not at either end.
So ,
You have 5 words,but one of them cannot be in either ends ,so you have 4 options for the first end.How many options do you have for the 2nd end ? 5 word - x = 4 options.But you already placed one of the options in the first end,so you have only 3 available words for the 2nd end.Since the 2nd spot in the row is not limited and you have already used 2 words out ot 5,you have 3 options for the 2nd spot and respectively 2 options for the 3rd and 1 for the 4th.
Now multiply the possible ways for each spot
4 ways for the first end x 3 ways for the second end x 3 ways for the second spot x 2 ways for the 3rd spot x 1 way for the 4th spot = 4x3x3x2x1 = 72 possible arrangements,if x cannot be in either end.</p>
<p>And i think i know why you keep getting 96. If you have 4 possible ways for the first end,you may have thought you have 4 possible ways for the second end as well.And when you multiplied 4x4x3x2x1 you answer was 96 ,instead you should have multiplied 4x3x3x2x1</p>
<p>An Alternate method is:
First the different combination with no restrictions is 5!=120
Now the total combination when x is in the first position is 4!=24
Similarly total combination with x in the last position is again 4!=24
Since x cannot be at either end, total no. of possible combinations is 120-(24+24)=72</p>