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In how many ways $P,Q,R,S,T,U$ can be arranged such that $ P, Q$ should come before $T,U$ ?

Do we have to find the ways that $P$ and $Q$ can be placed in the first four position?

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user2378
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  • Not necessarily, because $R,S$ might appear at any place. You'd want to pick an order in which $P,Q$ appear w.r.t. each other, then do the same with $T,U$. After that, you'd have to inject $R,S$ somewhere in there. – Jonathan Y. Oct 20 '13 at 18:17

1 Answers1

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We interpret the question as asking for the number of ways to arrange the letters so that $P$ is before $T$ and $U$, and also $Q$ is before $T$ and $U$.

Consider just the $4$ letters $P,Q,T,U$. There are $(2)(2)$ ways to arrange them so that our condition is fulfilled. And there are $4!$ ways to arrange them with no restriction.

Thus the probability that our arrangement of $6$ letters satisfies the condition is $\frac{4}{4!}$. There are $6!$ equally likely arrangements, so the required number of arrangements is $6!\cdot\frac{4}{4!}$.

Remark: We used a probabilistic argument. However, we can use pure counting. The positions to be occupied by our $4$ important letters can be chosen in $\binom{6}{4}$ ways. By our argument above, the actual locations of $P,Q,T,U$ can be chosen in $(2)(2)$ ways. Then the remaining $2$ letters can be put in the two open solots in $2!$ ways, for a total of $\binom{6}{4}(2)(2)(2!)$.

André Nicolas
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