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  1. The letters of the word CAMBRIDGE are arranged such that the first five letters of the alphabet are in alphabetical order. Find the number of different permutations. [ANS: 3024] Only permutation should be used.

  2. Two Families, each consisting of the father, the mother and a daughter, sit in a row. In how many ways can they be seated such that each arrangement is SYMMETRICAL with respect to (a) the 2 fathers, the 2 mothers and the 2 daughters. [ANS=6]

(b) the 2 men, the 2 women, and the 2 girls. [ANS=48]

I do not understand symmetrical arrangements. What is the difference between 2 fathers and 2 men. What are the arrangements for (a).

Thanks.

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    There is little to no difference between "2 fathers" and "2 men" in this problem, however there **is** a difference between "two mothers and two daughters" and "four women." In order to keep the language similar, they opted to refer to the men as fathers since they are distinguishing between the different categories of women. An answer of six would imply that the families with remain grouped so you have (Family1)(Family2) and the families are otherwise indistinguishable. (*I disagree with that interpretation and would have answered differently*) – 2017-02-16
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    A rewording of question 2a which might make more sense is "How many ways can you arrange the letters $FFMMDD$ into a palindrome (*so that it is read the same forwards as backwards*)." (*answer to this is $6$*). For 2b, "How many ways can you arrange the letters $\color{red}{FMD}\color{blue}{FMD}$ into a palindrome (*where for purposes of being considered a palindrome color doesn't matter, but for purposes of counting different arrangements color does matter*), the answer to this is $48$. I find the wording of the original question horrible and can see why you would be confused. – 2017-02-16

1 Answers 1

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$CAMBRIDGE$ is a $9$ letter word. Thus when constructing a permutation of this word, we have $9$ positions to work with. First reserve $5$ positions for $A,B,C,D,E$ and place these letters among those five positions so that they are in alphabetical order: there are exactly $\left( \begin{matrix} 9 \\ 5 \end{matrix} \right)$ ways to do this. For instance, here are a few ways

$XABXCXDCX$

$XXABCDEXX$

$ABCXXXDXE$,

where the $X$'s are filled in by the remaining letters $M,R,I,G$ in some fashion.

Next, note that after reserving positions for $A,B,C,D,E$, there are $4!$ ways to fill the remaining positions with the letters $M,R,I,G$. Thus the answer to the first question is

$4! \cdot \left( \begin{matrix} 9 \\ 5 \end{matrix} \right) = 3024$

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    Thanks Buddy ..Amazing!! I got the point. We must CHOOSE 5 slots out of 9. To do this, only combination is suitable and NOT permutation. – 2017-02-21