A 4-letter code word consists of letters A, B, and C. If the

Question

A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

A. 72
B. 48
C. 36
D. 24
E. 18

Bunuel · Accepted Answer

I don't see how the above way is giving 36 as an answer.

A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

A. 72
B. 48
C. 36
D. 24
E. 18

As the code must include all three letters, the pattern of the code word is  ABC - X , where X can be any letter out of A, B, or C. So we can have the code word consisting of letters:
 
ABC - A;
ABC - B;
ABC - C.
 
We can arrange the letters in each of the above 3 cases in  \frac{4!}{2!}  ways (since each case has 4 letters, with one letter repeated twice). So, the total number of code words is  3*\frac{4!}{2!} = 36 .

Answer: C.

Hope it helps.

gpulluri · Answer

It has to be 36 ..

We know that the code already contains the the letters A,B and C. Now the 4th letter can be choosen in 3 ways (A,B,C).

Once we have the set of 4 letters we can arrange them in 4!/2! ways.

Now the total ways of arranging would be 4!/2!*3 = 4*3*3 = 36

How about AACB ?

ANSWER:C

x2suresh · Answer

ABCA + ABCB + ABCC

= 4!/2! +4!/2!+4!/2!
= 36

walker · Answer

E

N=C^3_1*C^2_1*C^1_1*C^3_1=3*2*1*3=18

or

N=3^4-C^3_1-C^3_1*C^2_1*C^4_1-C^3_1*C^4_2*C^2_1*C^2_2=81-3*2*4-3*6*2*1=81-3-24-36=18

or

N=P^3_3*C^3_1=3*2*3=18

srp · Answer

E: 18

3 letters A, B and C can be arranged in 3! ways = 6 ways
4th letter can be chosen in 3 ways

Total number of ways = 6*3=18

CaspAreaGuy · Answer

but then you assume that 4th letter can stand only in the end, don't you?

srp · Answer

Right!

so we then get ABC  where R stands for repeat letter of type A,B,C

Number of ways of arranging ABC  = 4!/2! and since we have to do this 3 times for each A,B and C, Total ways = 4!/2! * 3 = 36

walker · Answer

if you think N=36, try to add a new 3-letters code to the 18-set:

ABCA,ABCB,ABCC
ACBA,ACBB,ACBC
BACA,BACB,BACC
BCAA,BCAB,BCAC
CABA,CABB,CABC
CBAA,CBAB,CBAC

walker · Answer

I corrected my formula:

C

N=\frac{C^3_3*C^3_1*P^4_4}{P^2_2}=\frac{1*3*4*3*2}{2}=36

or

N=3^4-C^3_1-C^3_1*C^2_1*C^4_1-\frac12 *C^3_1*C^2_1*C^4_2=81-3*2*4-\frac12 *3*6*2*1=81-3-24-18=36

urchin · Answer

36 for me too.
The group of 4 letters would be ABCX where X is A/B/C
just find out the permutation for 1 specific case say ABCA
4 things can be permuted in 4! ways and since 2 things are same(here 2 A's) divide by 2!
therefore... 4!/2!
Since there are three such groups based upon value of X , multiply by 3.
Ans: 3*(4!/2!) = 36

coldteleporter · Answer

(4!/2) + (4!/2) + (4!/2 ) = 3*(4!/2 ) = 36 .

Ans: C

Explanation:

Assuming A is the repeated letter, we get the 1st 4!/2,
 OR
if B is the repeated letter, we get the 2nd 4!/2
 OR
if C is the repeated letter, we get the 3rd 4!/2

that gives (4!/2) * 3 = 36

MBAwannabe10 · Answer

here is my approach: 
3*2*1*3 = 18 but there is 4!/2! of arranging them =>36 ways in the end

pavanpuneet · Answer

Here is how I tried to solve the question:

Consider XXXX = Assume the first three position is taken as for letters ABC those can be filled in 3! ways and then last letter can be filled in 3 ways... thus a total 18 ways.

Next, assume, that it XABC = 18 ways; next. CXAB = 18 ways; next BCXA = 18 ways... thus a total of 18*4 = 72ways!

Bunuel · Answer

Note that the correct answer to this question is 36, not 72.

A-ABC can be arranged in 4!/2!=12 ways;
B-ABC can be arranged in 4!/2!=12 ways;
C-ABC can be arranged in 4!/2!=12 ways;

Total: 12+12+12=36.

Answer: C.

Bunuel · Answer

We are not assuming that at all. The solution above says:

AABC can be arranged in 4!/2!=12 ways . So, for the case when we have 2 A's in a code, 12 different arrangements are possible:
AABC;
ABAC;
BAAC;
...

The same for other cases.

Hope it's clear.

SergeyOrshanskiy · Answer

First pick which letter is doubled. There are 3 ways. Without loss of generality, A is doubled.
Then pick a place for B. There are 4 ways to pick a place for B.
Then pick a place for C. There are 3 ways to pick a place for C.
Then place the two A's.

3*4*3=36.

It is similar to AKProdigy87's solution, except that we don't even need the 4C2. No formula whatsoever, just multiplication.

vad3tha · Answer

Could you please prove that "Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!" or show me the link that explain the formulation n!/r!?

Bunuel · Answer

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!} . Theory on Combinations: https://gmatclub.com/forum/math-combina ... 87345.html DS questions on Combinations: https://gmatclub.com/forum/search.php?s ... &tag_id=31 PS questions on Combinations: https://gmatclub.com/forum/search.php?s ... &tag_id=52 Tough and tricky questions on Combinations: https://gmatclub.com/forum/hardest-area ... 01361.html

happyface101 · Answer

Can someone please explain why 4!/2! represent the number of ways to arrange 4 letters of which one is repeated twice?? Thank you!

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