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1. In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
25,401,600
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
33,810
18 May 2010, 01:13
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chandrun
THEORY:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
Back to the original questions:
1. In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?
There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice.
1. Choosing 4 letters out of 10 (12-2(P and S)=10) to place between P and S = 10C4 = 210;
2. Permutation of the letters P ans S (PXXXXS or SXXXXP) = 2! =2;
3. Permutation of the 4 letters between P and S = 4! =24;
4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040;
5. We should divide multiplication of the above 4 numbers by 2! as there is repeated T.
Hence: \(\frac{10C4*2!*4!*7!}{2!}=25,401,600\)
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
There are 11 letters in the word "MISSISSIPPI ", out of which: M=1, I=4, S=4, P=2.
Total # of permutations is \(\frac{11!}{4!4!2!}\);
# of permutations with 4 I's together is \(\frac{8!}{4!2!}\). Consider 4 I's as one unit: {M}{S}{S}{S}{S}{P}{P}{IIII} - total 8 units, out of which {M}=1, {S}=4, {P}=2, {IIII}=1.
So # of permutations with 4 I's not come together is: \(\frac{11!}{4!4!2!}-\frac{8!}{4!2!}=33,810\).
Hope it helps.
03 Feb 2011, 03:42
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Bunuel
What about my approach to PERMUTATIONS?
The word permutations consists of 12 letters.
You can choose P and S in 7*2 ways so there are always 4 numbers between them
P on first, S on fifth ........ P on seventh S on twelwth + reversely (SP)
You are left with 10 letters, 2 of which are the same (TT)
So the complete formula is:
\(7*2*\frac{10!}{2!}=7*10!\)
It makes the same result, but I think is a bit quicker.
