IMO: 12

shankar.ashwin wrote:There are 4 pens each of a different color and four caps each of a diff color (But same as that of the pens). In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color.

A) 9
B) 12
C) 18
D) 23
E) 27

Pens: A, B, C, D <Colors depicted by alphabets>
Caps: a, b, c, d <small case letters depicting equivalent colors of the Pen >
Comboes allowed: Ab, Ac, Ad, Ba, Bc, Bd, Ca, Cb, Cd, Da, Db, Dc = 12
Ans > B

Proleefeek111 wrote: Pens: A, B, C, D <Colors depicted by alphabets>
Caps: a, b, c, d <small case letters depicting equivalent colors of the Pen >
Comboes allowed: Ab, Ac, Ad, Ba, Bc, Bd, Ca, Cb, Cd, Da, Db, Dc = 12
Ans > B

12 is incorrect.

Try listing down pairs as (Ab)(Bc)(Cd)(Da)
where ABCD are pens and abcd are caps.

The answer is A.. This is similar to that Tanya mailed 4 letters problem.
Look at the following link.

https://www.beatthegmat.com/difficult-pr ... 77390.html

That question is divided in two parts. The second part is the one relevant to this problem. Especially look at Anurag's explanation. And also one guy has listed the DeRangement formula which can help you find this answer in 15 seconds. lol

shankar.ashwin wrote:

Proleefeek111 wrote: Pens: A, B, C, D <Colors depicted by alphabets>
Caps: a, b, c, d <small case letters depicting equivalent colors of the Pen >
Comboes allowed: Ab, Ac, Ad, Ba, Bc, Bd, Ca, Cb, Cd, Da, Db, Dc = 12
Ans > B

12 is incorrect.

Try listing down pairs as (Ab)(Bc)(Cd)(Da)
where ABCD are pens and abcd are caps.

Shouldn't the problem statement be changed as follows then:

"In how many ways can these pens be arranged such that a pen and it's cap do not have the same color. "

as opposed to

"In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color. "

shankar.ashwin wrote:There are 4 pens each of a different color and four caps each of a diff color (But same as that of the pens). In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color.

A) 9
B) 12
C) 18
D) 23
E) 27

Number of ways for NO pen to receive the correct cap = total ways to arrange the 4 caps - number of ways for 1, 2 or all 4 pens each to receive the correct cap.

Note that if 3 pens each receive the correct cap, then ALL 4 pens each receive the correct cap.

Let the 4 pens = ABCD and the 4 caps = abcd.

Total number of ways to arrange the 4 caps = 4! = 24.

Exactly 1 pen receives the correct cap:
Let A be the one pen to receive the correct cap.
Number of options for A = 1. (Must be a)
Number of options for B = 2. (Must be c or d)
At this point, either C or D could still receive the correct cap.
Number of options for the pen that could still receive the correct cap = 1. (Must be the WRONG cap)
Number of options for the last pen = 1. (Must be the one remaining cap)
To combine these options, we multiply:
1*2*1*1 = 2.
Since any of the 4 pens could be the one to receive the correct cap, we multiply by 4:
2*4 = 8.

Exactly 2 pens each receive the correct cap:
Let A and B be the pair to receive the corrrect caps.
Number of options for A and B = 1. (A must receive a, B must receive b)
Number of options for C = 1. (Must be d)
Number of options for D = 1 . (Must be c)
To combine these options, we multiply:
1*1*1 = 1.
Since any combination of 2 pens could be the pair to receive the correct caps, we multiply by the number of combinations of 2 that can be formed from 4 choices:
1 * 4C2 = 6.

All 4 pens each receive the correct cap:
Number of options = 1.

Thus:
Number of ways for NO pen to receive the correct cap = 24-8-6-1 = 9.

The correct answer is A.

4 pens ABCD
4 cups abcd

A B C D

b a d c
b d a c
b c d a

c a d b
c d b a
c d a b

d a b c
d b a c
d c a c

Bek wrote:4 pens ABCD
4 cups abcd

A B C D

b a d c
b d a c
b c d a

c a d b
c d b a
c d a b

d a b c
d b a c --> B cannot couple B
d c a c--> A typo, i think u meant d c a b

Mayte, check the comboes .

is there a way for this to be solved by directly counting ways of allocating caps to pens(other than listing all combinations) rather than counting total and then subtracting unwanted cases?

IMO B.

1 Pen 3 other colors. So for each colored pen we have 3 combinations.
4 x 3 = 12 combinations totally

sohrabkalra wrote:is there a way for this to be solved by directly counting ways of allocating caps to pens(other than listing all combinations) rather than counting total and then subtracting unwanted cases?

Let the pens = ABCD and the caps = abcd.

Case 1: The pens are divided into pairs. Within each pair, the pens swap caps.
Number of options for A = 3. (Must receive b, c or d)
Number of options for the pen whose cap was given to A = 1. (Must receive a)
2 pens left. Each could still receive the correct cap, since the first 2 pens swapped caps.
Number of options for the next pen = 1. (Of the 2 remaining caps, must receive the wrong cap)
Number of options for the last pen = 1. (Only 1 cap left)
To combine these options, we multiply:
3*1*1*1 = 3.

Case 2: No two pens swap caps.
Number of options for A = 3. (Must receive b, c, or d)
Number of options for the pen whose cap was given to A = 2. (Of the 3 remaining caps, either of the 2 caps besides a)
2 pens left. Only one could still receive the correct cap, since the first two pens did not swap caps.
Number of options for the pen that could still get the correct cap = 1. (Must receive the wrong cap)
Number of options for the last pen = 1. (Only 1 cap left)
To combine these options, we multiply:
3*2*1*1 = 6.

Total ways = 3+6 = 9.