How many words, with or without meaning can be made from the

6c4 - what does this 'c' means?

c means "combinations"

\(nCk = C^n_k = \frac{n!}{(n-k)!k!}\)

Look at this post (Unfortunately, It's not finished yet): math-combinatorics-87345.html

sorry, I'm lost on step 4. Why is it 7 rather than 6?

thanks in advance.

Consider P, S and four letters between them as one unit: {PXXXXS}. 6 more letters are left, so total 7 units: {PXXXXS}, {X}, {X}, {X}, {X}, {X}, {X}. These seven units can be arranged in 7! # of ways.

Hope it's clear.

Bunuel, for question 1, why are we considering 4!?as the problem says, no letter should be repeated.but 4! would mean these 4 letters will repeat.pls help me understand

4! doesn't mean that letters are repeated. It means letters are re-arranged.

ABC can be re-arranged in 3!=3*2=6 ways.

ABC
ACB
BAC
BCA
CAB
CBA

If the letters were allowed to repeat: it would be 3*3*3=3^3=27 ways.

1.
6P4 = 6!/2!=6*5*4*3=360
OR
6C4*4!=360

If repetition were allowed, it would be (n^r)
6^4=6*6*6*6=1296 ways

I used an alternative approach.
\(P\)_ _ _ _ \(S\)_ _ _ _ _ _. The blank spots can be arranged with the remaining 10 letters in \(10!\) ways. The P and S can together be arranged in \(2!\) ways. Also since it is mentioned that have to be always 4 letters between P and S, hence this arrangement can be stretched to another 6 ways-all together 7 ways. Because of the repition of Ts, divide the entire relation by 2.
Hence,
\(10!*2*7 / 2\) or \(10!*7\).
Now I shall really appreciate, if anyone helps me in calculating this relation or that of Bunuel's quickly.

First of all you did everything correct: 10!*7=25,401,600. Next, this is not a GMAT question, because on the exam you won't be asked to calculate 10!*7. If it were a GMAT question, then most likely one of the options would be 10!*7, or we would be able to eliminate other options easily.

Solution 1:
6*5*4*3 = 360

Solution 2:
6!/(2!4!) * 4! = 6!/2! = 6*5*4*3 = 360 Using the Selection/Deselection Formula \(\frac{N!}{S!D!}\) then multiplying the selections by 4! to get the arrangement of the 4 letters.

More examples of the Selection/Deselection technique: Combinations: Deselection/Selection

Solution 1:
1. How many number of ways to position P _ _ _ _ S _ _ _ _ _ _ ? We can have P appear on the first, second, third, to..., seventh of the arrangement of letters. At the same time, we could have P _ _ _ _ S or S _ _ _ _ P.

Thus, 7 * 2!

2. How many ways to arrange 10 remaining letters? 10!/2! We divide by 2! because of 2 Ts in the PERMUTATIONS.

\(=\frac{7*2!*10!}{2!}=7*10!=25401600\)

Solution 2:
If you want to understand how permutations work in more detail...

How many number of ways to position P _ _ _ _ S _ _ _ _ _ _ ? We can have P appear on the first, second, third, to..., seventh of the arrangement of letters. At the same time, we could have P _ _ _ _ S or S _ _ _ _ P.

Thus, 7 * 2!

How many number of ways can we select four letters within the P_ _ _ _ S and outside? Getting the number of selections of those selected is always equal to that of those not selected. We use the Selection/Deselection Technique.

10!/4!6!

How many ways can we arrange the four letter within P and S? 4!
How many ways can we arrange the letters outside P and S? 6!
How many duplicate letters just 2 Ts? So we have to divide by 2!.

\(=\frac{7*2!*10!*4!*6!}{2!*4!*6!}=25401600\)

Hey thanks for the solution Bunuel, it is a silly doubt but I did not understand permutations of the 7 units, can you please help me? Why is it not permutations of 6 units?

There are 12 letters in "PERMUTATIONS". Four letters between P and S (total of six letters) is one unit: {P(S)XXXXS(P)}, the remaining 6 letters are also one unit each, so total of 7 units.

Hope it's clear.

It is now! Thanks

Bunnel ,
Need your help on this one.
I thought we could have the case like

YYYYYYPXXXXS
or
PXXXXSYYYYYY

but other combos are also possible like

YYYPXXXXSYYY

In the sense total sum of Ys has to be six but "how many are on P's side and how many on S's side is not fixed " am i missing something fundamental here ?

No, your understanding is correct.

2. In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S

A. 2419200
B. 25401600
C. 1814400
D. 1926300
E. 1321500

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: \(\frac{10C4*2!*4!*7!}{2!}=25,401,600\)

Answer: B.[/quote]


Hi Bunnel,

We have this -
- - - - - - P - - - - S , where dashes (-) can be arranged in whatever ways.
I took 3 cases :
1. When both T's will be within P and S , 2. - when both will outside of P and S and 3. - when one will be in and other out of Pa nd S.

I get below = ( 2*4!/2!*7! + 2*4!*7!/2! + 2*4!*7!) * 10C4

= 2*4!*7!*{1/2+1/2+1) * 10C4
= 2*4!*7!*2 * 10C4

Whats wrong in this calculation ? Please guide

If you put 2 T's between P and S, then multiplying this by 10C4 won't be correct. 10C4 is there the number of ways to choose 4 letter which will be between P and S and if already put 2 T's there then you choose 2 out of 8.

It should be \((2*\frac{4!}{2!}*7!*C^2_8+ 2*4!*\frac{7!}{2!}*C^4_8+ 2*4!*7!*C^3_8)\).

Those are non-GMAT questions. Locking the topic.