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I research some sites and books and i found these this approach helpful but could not understand a bit.

Approach: All the 7 letters of the word can be arranged in 7! ways.

and 3particular letters can be arranged in 3! ways..But the given condition is satisfied by one out of 6 ways. Here, i only know the 4 ways which are

1: A before I and I before O //only one way

2: A before I and I after O

3: A after I and I after O

4: A after I and I before O

the other 2 ways i couldn't figure out. eqn1

and said that the given condition is satisfied by 1 out of 6 ways Hence,the required no of ways=7!/3!=840 eqn2

My Question is in //1, //2

//1 the Other 2 ways which i couldn't figure out?

//2 Why we Divide by 3! in 7!/3!?

Jack
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    Order: AIO, AOI, IAO, IOA, OAI, OIA. 7! includes all these orders. You need 1 of them. – Bhaskar Aug 15 '15 at 06:22
  • @L16H7 Suppose if i need 2 of them then what will be the ans ?For example i need AIO and AOI – Jack Aug 15 '15 at 06:25
  • @JalajChawla elaborate on what you mean by that and why you would need 2 of them when the problem asks for A/I/O. – user217285 Aug 15 '15 at 06:29
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    If you were curious as to the number of arrangements that satisfy either of $A<I<O$ or $A<O<I$, then there would be two out of the six cases (AIO,AOI,IAO,IOA,OAI,OIA), so instead of dividing by six, you divide by three, since only a third of the cases are "good." Similarly, if following my answer below, it would be as though there was a step 1b, where you picked how the vowels were arranged in the three spaces such that they fit your requirement. – JMoravitz Aug 15 '15 at 06:34
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    Number 2 and number 4 have two variations, depending on the relationship of $A$ and $O$. – Arthur Aug 15 '15 at 06:34
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    Immediately before or anywhere before (i.e., can other letters come in between them)? – barak manos Aug 15 '15 at 06:39
  • Other letters can come in between A,I,O – Jack Aug 15 '15 at 06:43

2 Answers2

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An alternate approach would be to use combinations in addition to the traditional multiplication principle.

Step 1: Pick the three spaces used by the vowels $A, I, O$ simultaneously. There are $\binom{7}{3}$ ways to complete this step.

Now that the spaces have been picked, since we are only interested in those arrangements with $A$ before $I$ before $O$, we use those three spaces we picked to place $A$ in the earliest of the picked spaces, $I$ in the middle of the picked spaces, and $O$ in the final picked space.

Step 2-5: For each remaining space, pick one of the unused letters to occupy that spot. There are $4\cdot 3\cdot 2\cdot 1$ ways of completing these remaining steps.

By multiplication principle, there are then $\binom{7}{3}4!$ number of total arrangements with the property we sought.

Note: $\binom{7}{3}4!=\frac{7!}{3!}$

JMoravitz
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  • Okay +1 for this great Concept.But what is this eqn( 7/3) I haven't used it? – Jack Aug 15 '15 at 06:47
  • @Jalaj $\binom{n}{r}$, read as "$n$ choose $r$", is the binomial coefficient, the number of ways of choosing $r$ objects out of $n$ distinct objects. It has the form $\binom{n}{r}=\frac{n!}{r!(n-r)!}$. – JMoravitz Aug 15 '15 at 06:53
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The cases are {AIO,AOI,IAO,IOA,OAI,OIA}. The actual enumeration of the cases is irrelevant to the problem, so long as you know the number of ways they can be rearranged. The division by $3!$ exploits the symmetry in permutations. Suppose the set $X_1$ contains all the permutations of RAINBOW with A before I before O, $X_2$ contains all the permutations of RAINBOW with A before O before I, etc. all the way to $X_6$, which contains the permutations with O before I before A. The latter 5 sets can be obtained from the first set switching the orders of the A/O/I as prescribed. In this way, all the sets have the same number of elements, with the union containing all the permutations. Thus the number we seek is $\frac{7!}{3!}$. In a much larger problem where we can't list out the permutations as done in the first line, apply the same reasoning and recognize that the permutations can be partitioned evenly, with exactly one of those subsets satisfying the criterion.

user217285
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  • I could not understand from here"The latter 5 sets can be obtained from the first set switching the orders of the A/O/I as prescribed. In this way, all the sets have the same number of elements, with the union containing all the permutations. Thus the number we seek is 7!3!......(till the end)"Can u explain me with any real world example so that i can relate more why they divided 7!/3! – Jack Aug 15 '15 at 06:43