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14 Aug 2020, 00:45
Kudos
Bookmarks
arun6765
Hi Bunuel
I have also followed the same approach - 8C4 * (7!/4!2!) (8C4 for filling 4 I's in 8 blank spaces and arranging rest 7 alphabets in 7!/4!2!)
The answer is not matching with the explanation given to the question.
The steps i followed seems logical, so it will be very helpful if you can point out the gap in my understanding.
Thanks in advance
29 Jan 2021, 17:00
Kudos
Bookmarks
This is a tough question.
Here's how I thought about it. Can someone chime in on whether the approach is correct?
P_ _ _ _ S _ _ _ _ _ _
S_ _ _ _ P _ _ _ _ _ _ <--- Put P and S like so in either of these two configurations
4 letters in between P and S can follow one of these two options:
a) Both Ts are included: 10! / 2! x 6! /(6 - 6)! <--- First part is with the 2! indicating T repeats and second part is choosing 6 letters for the last spots after S or P
b) Both Ts are not included : 10! /(10 - 4)! x 6! / 2! <--- Reverse of option a)
Add a) and b) together then multiple by 2 since P and S can be arranged in two days with respect to each other.
Here's how I thought about it. Can someone chime in on whether the approach is correct?
P_ _ _ _ S _ _ _ _ _ _
S_ _ _ _ P _ _ _ _ _ _ <--- Put P and S like so in either of these two configurations
4 letters in between P and S can follow one of these two options:
a) Both Ts are included: 10! / 2! x 6! /(6 - 6)! <--- First part is with the 2! indicating T repeats and second part is choosing 6 letters for the last spots after S or P
b) Both Ts are not included : 10! /(10 - 4)! x 6! / 2! <--- Reverse of option a)
Add a) and b) together then multiple by 2 since P and S can be arranged in two days with respect to each other.
23 Apr 2026, 13:12
Kudos
Bookmarks
1. PERMUTATIONS there are always 4 letters between P & S
P[4 letters]S and this lump can be placed in the 7 gaps created by arrangement of the leftover 6 letters.
=(10C4*!4) *(!6 * 7)/!2
[ here !2---for the two T's & leaving P&S, 4 letters can be selected and distributed in 10C4*!4]
since P&S can be arranged among themselves in two ways
so =(10C4*!4) *(!6 * 7)=(!10*!4*!6*7)/(!6*!4)
=!10*7=25,401,600
2. MISSISSIPPI where the 4 I's never come together
this is the simplest case: total arrangements of the given word MINUS the arrangements where the 4 I's are together
=!11/(!4*!4*!2) - !8/(!4*!2)
=!8/(!4*!2)[(11*10*9)/24 - 1]
=(8*7*6*5*161)/8=33810
P[4 letters]S and this lump can be placed in the 7 gaps created by arrangement of the leftover 6 letters.
=(10C4*!4) *(!6 * 7)/!2
[ here !2---for the two T's & leaving P&S, 4 letters can be selected and distributed in 10C4*!4]
since P&S can be arranged among themselves in two ways
so =(10C4*!4) *(!6 * 7)=(!10*!4*!6*7)/(!6*!4)
=!10*7=25,401,600
2. MISSISSIPPI where the 4 I's never come together
this is the simplest case: total arrangements of the given word MINUS the arrangements where the 4 I's are together
=!11/(!4*!4*!2) - !8/(!4*!2)
=!8/(!4*!2)[(11*10*9)/24 - 1]
=(8*7*6*5*161)/8=33810
