An English-language class has 8 students: one man and one

Question

An English-language class has 8 students: one man and one woman from each of four countries. The students are to be divided into 4 pairs so that no pair contains two students of the same gender or from the same country. How many distinct pairings are possible?

A. 4
B. 6
C. 7
D. 8
E. 9

A. 4
B. 6
C. 7
D. 8
E. 9

rajdeep212003 · Answer

I tried approaching this by thinking in terms of pairing men and women from different countries, but I’m still not seeing a clean or fast way to count the total distinct pairings without getting messy.

Could you please explain the shortest or most efficient way to think about this problem? I feel like there’s probably a neat pattern or setup here that I’m missing.

Would really appreciate a structured breakdown of the logic.

— Rajdeep

stne · Answer

c_1: m_1, w_1 
 c_2: m_2,w_2 
 c_3: m_3,w_3 
 c_4: m_4,w_4

m_1*3 = 3  ( After choosing  m_1  we have  3  options for woman)
 m_2*3 = 3  ( After choosing  m_2  we have  3  options for woman) 
 m_3*1 = 1  ( After choosing  m_3  we have only  1  option for woman) 
 m_4*1 = 1  ( After choosing  m_4  we have only  1  option for woman)

Total ways :  3*3*1*1= 9

Ans E

Hope it helped.

atharvadixit · Answer

what is m1 chooses w3, then m2 would have 2 choices, m3 would have 2 choices and m4 would have 1, therefore 3*2*2*1 =12???

stne · Answer

You are correct. Thank you for pointing this out.

This method does not work out in all the cases. Seems we have to use inclusion-exclusion or some other method to solve this.
Unfortunately I am not very well versed in that method.

Let's hope some one posts a good solution soon.

KarishmaB

KarishmaB · Answer

This question is the same as the standard Letters and Envelopes questions in Combinations discussed in GMAT curricula: 
In how many ways can we put 4 letters in 4 addressed envelopes such that no letter goes into its correct envelope?
The constraints are exactly the same. We cannot put a letter in a letter (so no man - man pairs) and we cannot put a letter in its correct envelope so no man-woman pair of the same country.

Ma can be paired in 3 ways (Wb, Wc, Wd). Say it is paired with Wc. 
Now we have 3 men and 3 women (Mb, Mc, Md and Wa, Wb, Wd). Now how many options do we have for Mc (since Wc is already paired)? 3 options. Mc can be paired with any of the available 3 women.

Say Mc were paired with Wd. Now we are left with Mb, Md and Wa, Wb. There is only 1 way to pair them because Mb-Wb is not allowed. 
If Mc were paired with Wa, then leftover people would be Mb, Md and Wb, Wd. Still only 1 way to pair them Mb-Wd and Md-Wb.

Hence number of combinations = 3*3 = 9

Answer (E)

stne · Answer

Karishma,  atharvadixit  had posted this doubt

(Ma,Mb,Mc,md ) (Wa,Wb,Wc,Wd)

Ma has three choices, suppose Ma chooses Wc.

Now Mb will have only two choices Wa and Wd, suppose he chose Wa.

Now Mc will have only two choices Wb and Wd, let's suppose he chose Wd.

Finally, Md will have only one choice,Wc.

Total count =  3*2*2*1 = 12

What is wrong with this? Thank you.

Gun99 · Answer

Hi Stne,

In your case you are voilating the same country rule and so answer 12 will be wrong here. For example -

Lets consider your scenario -
M1 takes W3 
Then M2 has 2 choices W1 or W4
but M3 has only 1 choice after this not 2. Suppose M2 selects W1 then M3 needs to select W4 else we voilate the same country rule, and suppose M2 selects W4, then M3 has 2 choices W1 and W2 and finally M4 has one.
So total 3 cases when we select W3.

in numbers explanation -

when M1 has taken W3 -
Case 1 M2 takes W1 -
M1 has 1 choice W3, M2 has 1 choice W1, then M3 has 1 choice only W4 and M4 has W2.

We get only 1 case in this.

Case 2 M2 takes W2 -
M1 has 1 choice W3, M2 has 1 choice W4, now M3 has 2 choices and then W4 has 1.
We get 2 cases in this case.

So total 3 cases form when we select W3 and doesn't voilate any rule, similarly it will be for W2 and W4 and hence,
the total cases will 3+3+3 = 9.

Hope this helps.

KarishmaB · Answer

When you say "suppose he chooses X out of n possible choices," ensure that the further choices are not path dependent.

(Ma,Mb,Mc,Md ) (Wa,Wb,Wc,Wd)

Ma has three choices (fine since Wb, Wc and Wd are equivalent for now). Say he chose Wc. 
Now Mb will have only two choices Wa and Wd. You must take both paths separately.

So you say 
- "if Mb chooses Wa, then Mc must choose Wd so that Md gets Wb". Note that here Mc does not have 2 choices since Md cannot be left with Wd. 
- But "if Mb chooses Wd, then Mc has 2 choices: Wa and Wb"

Now, what if instead Ma had chosen Wb? Then this whole case goes for a toss because Mb would have 3 options.

So instead of all this, after Ma chooses Wc, you need to go to Mc, the partner of Wc. For Mc, Wc is missing because of which further choices are not path dependent.

See the difference?

An English-language class has 8 students: one man and one

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