Number of ways in which the letters of the word RAINBOW be arranged such that n and b are together
$\begingroup$ The given question:
I gave it a try and thought below:
But the given answer is just 840 with following explanation:
So what's wrong with my logic? I must be counting certain arrangements multiple times or my logic must be at fault in large. But what's that exactly?
asked Apr 30, 2015 at 21:28
$\endgroup$ 3 $\begingroup$ For the arrangement RAINBOW you have counted four places the R can be, four places the W can be, and six ways of putting NB in that order between the I and O, so you have counted it $96$ times. Other arrangements have been counted multiple times as well. answered Apr 30, 2015 at 21:33
Ross MillikanRoss Millikan 364k27 gold badges244 silver badges433 bronze badges $\endgroup$ $\begingroup$ Ross Millikan has already explained the error in your reasoning. Here’s a correct version of the kind of calculation that you were trying to make. There are $\binom73$ ways to choose $3$ positions for the letters A, I, and O. The remaining $4$ letters can fill the remaining $4$ slots in any order, so they can be inserted in $4!$ ways. That gives you a total of $$4!\binom73=\frac{7!}{3!}=840$$ arrangements. answered Apr 30, 2015 at 21:36
Brian M. ScottBrian M. Scott 592k53 gold badges717 silver badges1182 bronze badges $\endgroup$ $\begingroup$ Your mistake is here:
The 43680 placements of letters into four of the 16 positions don’t yield different arrangements of the letters. For example, one arrangement is RAINBOW. Is the letter R in position 1, 2, 3, or 4? You counted each of those placements separately. answered Apr 30, 2015 at 21:35
Steve KassSteve Kass 14.3k1 gold badge19 silver badges32 bronze badges $\endgroup$ $\begingroup$ You would be entirely correct if there really were 16 places ( 19 in all ) and the places not filled with letters were filled with spaces so that R***A****I*N*BO**W*was truly distinct from *R**A****I*N*BO***W ( I has used * because spaces are hard to count ) what $ ^{16} P _4$ counts is the number of four-tuples with non-repeating elements from 1 to 16. You might have the first element represent the position ( in your scheme of 16 positions ) of the letter "R" and the other three the positions of "N", "B", and "W" respectively. so that R***A****I*N*BO**W* would be represented as (1, 10, 12, 15) and *R**A****I*N*BO***W would be represented as (2, 10, 12, 16) . answered Apr 30, 2015 at 23:03
WW1WW1 9,7571 gold badge15 silver badges15 bronze badges $\endgroup$ $\begingroup$ I think the closest method to your original attempt is the following: after laying out the fixed letters A, I and O, consider one additional letter at a time. (1) We have: " A I O ". R can be placed in any of four places. (2) We now have four placed letters, for example " R A I O ". In any case we have five locations in which the next letter N can be placed. (3) Similarly when we place the B, there are 6 choices of where to put it among the existing 5 letters. (4) And finally the last letter W can be placed among the 6 existing letters in any of 7 places. Each combination of choices will lead to a different outcome, as any two outcomes must differ by the order of as at least one pair of letters. So the total number of orders is: 4 * 5 * 6 * 7 = 840 answered May 1, 2015 at 8:28
IanF1IanF1 1,4691 gold badge11 silver badges13 bronze badges $\endgroup$ $\begingroup$ I got the answer above, but when I tried to solve the same problem I applied slightly different thinking to same solution. Thus sharing it here. The book says:
We can also think it as follows:
Just a little bit of different thinking. answered Dec 23, 2015 at 11:33
MahaMaha 1,8932 gold badges26 silver badges41 bronze badges $\endgroup$ |