How many 5 letter words ( with or without meaning) can be formed using

Question

How many 5 letter words ( with or without meaning) can be formed using all the following 5 letters P,Q,R,S,and T so that letter P is to the left of letter R?

(A) 120

(B) 60

(C) 48

(D) 24

(E) 12

GMATinsight · Accepted Answer

Best Method:

Total Ways of arranging 5 letters in any possible order = 5*4*3*2*1 = 5! = 120

in Half of the cases P will be to the left of R and in other half P will be to the right of R

Hence, Desired outcome = 120/2 = 60

Alternate method:

we have five blank spaces _ _ _ _ _ which have to occupied by these 5 letter such that P is to the left of R

so let's select two blank spaces for P and R out of 5 blank spaces,

No. of ways of selecting two blank spaces out of 5 = 5C2 = 10

The no. of ways of arranging P and R on selected two places such that P is to the left of R = 1

No. of ways of arranging remaining three letters on remaining three blank spaces = 3*2*1 = 3! = 6

Total Ways of Arrangement = 5C2 * 1 * 3! = 10 * 1 * 6 = 60

Answer: option B

Skywalker18 · Answer

It seems there are only 4 distinct letters in the question stem .

Considering that there are 5 distinct letters - P,Q,R,S,T Number of words - 5!/2 = 120/2=60 Using symmetry , in half of the 120 cases , P will be to the left of R while in the other 60 cases letter P will be to the right of R .

ENGRTOMBA2018 · Answer

We can solve this qustion by 2 methods,

Method 1:  Total combinations for 5 letters = 5! = 120

As there is no bias in counting the combinations, half of the combinations will have R to the left of P and half of them to the right of P.

Thus possible combinations = 120/2 = 60. B is the correct answer.

Method 2:

The combinations possible are:

PRQST, combinations of RQST = 4! =24
QPRST, combinations = 3C1*3!, 3C1 taken to account for the fact that instead of Q we can also take S or T
QSPRT, combinations = 2!*2!*3C2, 3C2 taken to account for the fact that instead of QS we can also take ST or QT, 2! each to take into account permutations for QS and RT
QSTPR, combinations = 3!*1, 3! to account for permutations for QST

Thus, the total combinations possible = 4!+3C1*3!+2!*2!*3C2+3! = 24+18+12+6=60. B is the correct answer.

AryamaDuttaSaikia · Answer

Conceptual way

As a general strategy in basic Question on Permutation Combination like this we

first arrange the letters on which there is a condition and then arrange the

remaining letters.

So out of the five positions where the 5 letters needs to be arranged we will first

arrange letters P and R.

So letters P and R will take two positions so lets select two positions out of 5

positions where letters P and R will be placed

Two positions can be selected out of 5 positions in 5C2 ways = 10 ways.

Once two positions have been selected we know P will be in the left position and

R will be in right so there is only one way we can place letters P and R in these

two positions.

And the remaining 3 letters can be placed in 3 posions in 3! Ways = 6 ways.

So the final Answer = number of ways letters P and R can be arranged x number

of ways the other 3 letters can be arranged

Final Answer = 10 x 6 = 60 ways

JAMBOREE Way

Now 5 letters can be placed in 5 positions in 5! Ways = 120.

Logically we can analyze that either letter P will be to the left of letter R or to the

right of letter R.

So by simple reasoning we will get that in half of the ways letter P will be to the

left of letter R.

So the correct answer = ½ total number of ways

= 60 ways

We make the  GMAT  simple!

22gmat · Answer

Why can't I consider PR as one letter? So I only have 4 letters which can be rearranged. (PR) Q S T n = 4! = 24? Where is my mistake? I don't understand it

Thank you very much.

Skywalker18 · Answer

You have only considered the case in which R is to the immediate right of P , but as per the question stem R can take any of the positions to right of P ( other letters can come in between) P _ _ _ _ R can take any of the 4 positions as long as it satisfies being to right of P. PRQST , Combination of RQST =4!=24 whereas as per your procedure , (PR) Q S T , there will be 3! combinations for this configuration. And you can proceed similarly for the other configurations . QPRST-3C1*3! QSPRT- 2!*2!*3C2 QSTPR- 3!*1 Hope it helped !!

KarishmaB · Answer

You might find this post useful: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... s-part-ii/

22gmat · Answer

Thank you all! Got it

yashkhandelwal25 · Answer

P and R: 5C2=10 choices 
 The remaining 3 letters can be arranged in 3!=6 ways. Total =10×6=60.

How many 5 letter words ( with or without meaning) can be formed using

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