I am getting a bit stuck on the theory behind the 4 I's calculation. Could you explain in more detail why it is 8!/4!2!?
chandrun wrote:
Here a couple of problems I encountered. Let me know your views on them. They may not necessarily be of GMAT format. Please share your views, nevertheless.
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?
Cheers
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.