[wolfgang]

There are three red Matryoshka dolls on the table.These dolls are of decreasing sizes placed inside each other.the total number of the dolls is 10.

3 Large(Mothers)all are red in color.

3 Middle(Daughters),they are either red or green .

3 Small(Children),they are either red or green.

1 Green baby doll(this baby can be inside anyone of the three sets).

you can open only one Large mother Matryoshka, to decide where is the baby(i.e. in which set?).

these informations are given:

1- only one set has (2 red dolls) under each other.

2- only one set has (2 green dolls) under each other.

[Guest]

Question: Can you open the dolls underneath the large mother matryoshka you choose, or only open that one?

[wolfgang]

Question: Can you open the dolls underneath the large mother matryoshka you choose, or only open that one?

You can open only the mother.

[araver]

First, I assume you won't accept any attempt of weighing the dolls - I am assuming the color does not influence weigth as much as an extra baby doll. So, I dismissed this thought.

I had 2 questions regarding wording. Here are the two questions:

Q1) "under each other" is a bit confusing since they can't be both under each other. "Each other" is a reciprocal wording. A and B are in relation x with each other means both AxB and BxA. (Same problem in the second statement "These dolls are of decreasing sizes placed inside each other" means "each placed inside another". )

I presume you mean adjacent i.e. one inside the other, directly touching?

Q2) When you say "only one set has (X color dolls)" do you mean exactly X or at least X?

Then I realized that no possible combination of answers leads to a puzzle with a solution.

There are 8 Mother, Daughter, Children, Baby/No Baby combinations possible:

1 R R R

2 R R R G

3 R R G 

4 R R G G

5 R G R

6 R G R G

7 R G G 

8 R G G G

Assume answer for Q1 is "adjacent" and answer for Q2 is "exactly" yields no clear solution:

Info 1) Only combinations 3 and 4 have exactly 2 Red adjacent dolls. Since there is only 1 set with exactly 2 Red adjacent dolls, this means there is exactly one of combinations 3 or 4 in the three sets.

Info 2) Only combinations 4 and 7 have exactly 2 Green adjacent dolls. Since there is only 1 set with exactly 2 Green adjacent dolls, this means there is exactly one of combinations 4 or 7 in the three sets.

Assume combination 3 is in the three sets, then combination 4 cannot be (from Info 1). Therefore combination 7 must surely be in the sets (from Info 2). Since both 3 and 7 don't have the baby, there must be either 2 or 6 or 8 as the last set. However, opening one Mother means seeing only the Daughter's color.

So if you open one Mother and see Red, you might be looking at combination 3 or combination 2. No way of knowing for sure.

Also if you open one Mother and see Green, you might be looking at combination 7 or combination 8. No way of knowing for sure.

Assume combination 4 is in the three sets, then combinations 3 and 7 cannot be. Since combination 4 has the baby, combinations 2,6,8 are also not possible since they would contain a baby also. So the other sets have combinations 1 and 5. Again, assume you open a Mother and see Red. You might be looking at combination 1 (RRR) or at combination 4 (RRGG). No way of knowing. If both sets which don't contain the baby are the same combination (nothing in the wording of the puzzle excludes this case) then imagine they're both combination 5. Now if you open a Mother and see Green, then your looking at one possible combination 5. No way to know which of the other two is combination 5 and which is combination 4.

Assume answer for Q1 is "adjacent" and answer for Q2 is "at least" yields no clear solution:

Info 1) Only combinations 1,2,3 and 4 have at least 2 Red adjacent dolls. Since there is only 1 set with at least 2 Red adjacent dolls, this means there is exactly one of combinations 1, 2, 3 or 4 in the three sets.

Info 2) Only combinations 4, 7 and 8 have at least 2 Green adjacent dolls. Since there is only 1 set with at least 2 Green adjacent dolls, this means there is exactly one of combinations 4,7 or 8 in the three sets.

Assume combination 4 is in the sets. Then combinations 1,2,3,7,8 cannot be from Info 1 and 2. 6 cannot be since 4 already contains a baby. Which leaves only 2 copies of 5 possible. But in this case, if you open a Mother and see Green, you just discovered a combination of 5 and have no idea which of the remaining two is combination 4.

Assume combination 8 is in the sets. Then combinations 2,4,6 aren't (since 8 contains a baby), 7 isn't (from info 2). From info 1 you get exactly one of combinations 1 and 3 are in the three sets. So the third set is surely combination 5. So, if you open a Mother and see Green you're either looking at 5 or 8. No way of knowing.

If neither 4 nor 8 are in the sets, then 7 is in exactly one of the sets(According to info 2). From Info 1 exactly one of 1, 2 and 3 is. If 2 is, then the third set is not 6 so it's 5. Again, if you open a Mother and see Green you're either looking at 5 or 7, no way of knowing.. If 2 is not, then exactly one of 1 and 3 is. The baby is then in the third set, which is not 4, nor 8, nor 2 therefore is 6. Again, if you open a Mother and see Green you're either looking at 6 or at 7, no way of knowing.

Assume answer for Q1 is "not necessarily adjacent (simply inside)", and answer for Q2 is "exactly" yields no clear solution:

Info 1) Only combinations 3, 4, 5 and 6 have exactly 2 Red dolls. Since there is only 1 set with exactly 2 Red dolls, this means there is exactly one of combinations 3, 4, 5 and 6 in the three sets.

Info 2) Only combinations 4,6 and 7 have exactly 2 Green dolls. Since there is only 1 set with exactly 2 Green adjacent dolls, this means there is exactly one of combinations 4,6 and 7 in the three sets.

Assume that combination 4 is in the sets. Then from Info 1 and Info 2 combinations 3,5,6,7 are not possible. 2 and 8 are also not possible since 4 contains the baby. That leaves only combination 1 possible. So, opening a Mother will always show you Red and be useless in determining combination 1.

Assume that combination 6 is in the sets. Then from Info 1 and Info 2 combinations 3,4,5,7 are not possible. Again 2 and 8 are also not possible since 6 already contains the baby, which leaves combination 1 (2 copies). So, if you open a Mother and see Red, you've discover one of combination 1 copies, but still don't know which of the other has the baby.

If both combination 4 and combination 6 are not in the sets then from info 2 combination 7 is. Also, from info 1 exactly one of combinations 3 and 5 are. Again, if combinations 5 and 7 are in the sets, then opening a Mother and seeing Green just tells you you're looking at either 5 or 7 and can't help you find the baby. If combinations 3 and 7 are in the set, then the third is either 2 or 8 (since the third contains the baby and isn't 4 or 6). Again, looking at a Mother and seeing Red means you're either looking at 3 or 2 and seeing Green means you're either looking at 7 or 8. No way to know for sure.

Assume answer for Q1 is "not necessarily adjacent (simply inside)", and answer for Q2 is "at least" yields no clear solution:

Info 1) Only combinations 1,2, 3, 4, 5 and 6 have at least 2 Red dolls. Since there is only 1 set with at least 2 Red dolls, this means there is exactly one of combinations 1,2, 3, 4, 5 and 6 in the three sets.

Info 2) Only combinations 4,6,7 and 8 have at least 2 Green dolls. Since there is only 1 set with at least 2 Green adjacent dolls, this means there is exactly one of combinations 4,6,7 and 8 in the three sets.

Info 1 says that the other 2 of the sets have combinations 7 or 8. Since they both cannot have 8 (because it contains the doll) and they cannot be one 7 and the either 8 (because of Info 2), this means both sets are 7. Which means the set in Info 1 has the baby (2, 4 or 6).

If it's 2 or 4 and you open one Mother and see Green, you've lost since you don't know which of the others is 7 and which has the doll.

If it's 6, then opening any Mother will show you Green and you're lost.

[wolfgang]

First, I assume you won't accept any attempt of weighing the dolls - I am assuming the color does not influence weigth as much as an extra baby doll. So, I dismissed this thought.

I had 2 questions regarding wording. Here are the two questions:

Q1) "under each other" is a bit confusing since they can't be both under each other. "Each other" is a reciprocal wording. A and B are in relation x with each other means both AxB and BxA. (Same problem in the second statement "These dolls are of decreasing sizes placed inside each other" means "each placed inside another". )

I presume you mean adjacent i.e. one inside the other, directly touching?

Q2) When you say "only one set has (X color dolls)" do you mean exactly X or at least X?

Then I realized that no possible combination of answers leads to a puzzle with a solution.

There are 8 Mother, Daughter, Children, Baby/No Baby combinations possible:

1 R R R

2 R R R G

3 R R G 

4 R R G G

5 R G R

6 R G R G

7 R G G 

8 R G G G

Assume answer for Q1 is "adjacent" and answer for Q2 is "exactly" yields no clear solution:

Info 1) Only combinations 3 and 4 have exactly 2 Red adjacent dolls. Since there is only 1 set with exactly 2 Red adjacent dolls, this means there is exactly one of combinations 3 or 4 in the three sets.

Info 2) Only combinations 4 and 7 have exactly 2 Green adjacent dolls. Since there is only 1 set with exactly 2 Green adjacent dolls, this means there is exactly one of combinations 4 or 7 in the three sets.

Assume combination 3 is in the three sets, then combination 4 cannot be (from Info 1). Therefore combination 7 must surely be in the sets (from Info 2). Since both 3 and 7 don't have the baby, there must be either 2 or 6 or 8 as the last set. However, opening one Mother means seeing only the Daughter's color.

So if you open one Mother and see Red, you might be looking at combination 3 or combination 2. No way of knowing for sure.

Also if you open one Mother and see Green, you might be looking at combination 7 or combination 8. No way of knowing for sure.

Assume combination 4 is in the three sets, then combinations 3 and 7 cannot be. Since combination 4 has the baby, combinations 2,6,8 are also not possible since they would contain a baby also. So the other sets have combinations 1 and 5. Again, assume you open a Mother and see Red. You might be looking at combination 1 (RRR) or at combination 4 (RRGG). No way of knowing. If both sets which don't contain the baby are the same combination (nothing in the wording of the puzzle excludes this case) then imagine they're both combination 5. Now if you open a Mother and see Green, then your looking at one possible combination 5. No way to know which of the other two is combination 5 and which is combination 4.

Assume answer for Q1 is "adjacent" and answer for Q2 is "at least" yields no clear solution:

Info 1) Only combinations 1,2,3 and 4 have at least 2 Red adjacent dolls. Since there is only 1 set with at least 2 Red adjacent dolls, this means there is exactly one of combinations 1, 2, 3 or 4 in the three sets.

Info 2) Only combinations 4, 7 and 8 have at least 2 Green adjacent dolls. Since there is only 1 set with at least 2 Green adjacent dolls, this means there is exactly one of combinations 4,7 or 8 in the three sets.

Assume combination 4 is in the sets. Then combinations 1,2,3,7,8 cannot be from Info 1 and 2. 6 cannot be since 4 already contains a baby. Which leaves only 2 copies of 5 possible. But in this case, if you open a Mother and see Green, you just discovered a combination of 5 and have no idea which of the remaining two is combination 4.

Assume combination 8 is in the sets. Then combinations 2,4,6 aren't (since 8 contains a baby), 7 isn't (from info 2). From info 1 you get exactly one of combinations 1 and 3 are in the three sets. So the third set is surely combination 5. So, if you open a Mother and see Green you're either looking at 5 or 8. No way of knowing.

If neither 4 nor 8 are in the sets, then 7 is in exactly one of the sets(According to info 2). From Info 1 exactly one of 1, 2 and 3 is. If 2 is, then the third set is not 6 so it's 5. Again, if you open a Mother and see Green you're either looking at 5 or 7, no way of knowing.. If 2 is not, then exactly one of 1 and 3 is. The baby is then in the third set, which is not 4, nor 8, nor 2 therefore is 6. Again, if you open a Mother and see Green you're either looking at 6 or at 7, no way of knowing.

Assume answer for Q1 is "not necessarily adjacent (simply inside)", and answer for Q2 is "exactly" yields no clear solution:

Info 1) Only combinations 3, 4, 5 and 6 have exactly 2 Red dolls. Since there is only 1 set with exactly 2 Red dolls, this means there is exactly one of combinations 3, 4, 5 and 6 in the three sets.

Info 2) Only combinations 4,6 and 7 have exactly 2 Green dolls. Since there is only 1 set with exactly 2 Green adjacent dolls, this means there is exactly one of combinations 4,6 and 7 in the three sets.

Assume that combination 4 is in the sets. Then from Info 1 and Info 2 combinations 3,5,6,7 are not possible. 2 and 8 are also not possible since 4 contains the baby. That leaves only combination 1 possible. So, opening a Mother will always show you Red and be useless in determining combination 1.

Assume that combination 6 is in the sets. Then from Info 1 and Info 2 combinations 3,4,5,7 are not possible. Again 2 and 8 are also not possible since 6 already contains the baby, which leaves combination 1 (2 copies). So, if you open a Mother and see Red, you've discover one of combination 1 copies, but still don't know which of the other has the baby.

If both combination 4 and combination 6 are not in the sets then from info 2 combination 7 is. Also, from info 1 exactly one of combinations 3 and 5 are. Again, if combinations 5 and 7 are in the sets, then opening a Mother and seeing Green just tells you you're looking at either 5 or 7 and can't help you find the baby. If combinations 3 and 7 are in the set, then the third is either 2 or 8 (since the third contains the baby and isn't 4 or 6). Again, looking at a Mother and seeing Red means you're either looking at 3 or 2 and seeing Green means you're either looking at 7 or 8. No way to know for sure.

Assume answer for Q1 is "not necessarily adjacent (simply inside)", and answer for Q2 is "at least" yields no clear solution:

Info 1) Only combinations 1,2, 3, 4, 5 and 6 have at least 2 Red dolls. Since there is only 1 set with at least 2 Red dolls, this means there is exactly one of combinations 1,2, 3, 4, 5 and 6 in the three sets.

Info 2) Only combinations 4,6,7 and 8 have at least 2 Green dolls. Since there is only 1 set with at least 2 Green adjacent dolls, this means there is exactly one of combinations 4,6,7 and 8 in the three sets.

Info 1 says that the other 2 of the sets have combinations 7 or 8. Since they both cannot have 8 (because it contains the doll) and they cannot be one 7 and the either 8 (because of Info 2), this means both sets are 7. Which means the set in Info 1 has the baby (2, 4 or 6).

If it's 2 or 4 and you open one Mother and see Green, you've lost since you don't know which of the others is 7 and which has the doll.

If it's 6, then opening any Mother will show you Green and you're lost.

WooooooW!!very interesting attempts , I said they are in decreasing sizes,and one inside the other,and logically the big one(mother)will contain the(daugher)and the daughter will contain the (child),and only one set has an additional

(baby) inside the(child).

In your 8 combinations:

No. 1 & 2 are wrong...because there are three red colored dolls under each other.

No. 4 is also wrong...because it has both two red and two green colored dolls.

No. 8 is wrong ...because it has 3 greens successively.

So the only right answer(that was in my head)is:

3,6 and 7

and the baby is in No. 6

Thank you.....you are GREAT!!!

[fabpig]

Am I missing something really obvious?

Knowing that the answer is combination "6" in the example doesn't really help:

Opening mother to reveal another red means that the baby MUST be in one of the other sets (as it's the only set with 2 consecutive reds) - no way of knowing which one.

Opening mother to reveal green means the baby MAY be in this set, but equally could be in the other set which starts Red Green - again no way of knowing which one.

[wolfgang]

Am I missing something really obvious?

Knowing that the answer is combination "6" in the example doesn't really help:

Opening mother to reveal another red means that the baby MUST be in one of the other sets (as it's the only set with 2 consecutive reds) - no way of knowing which one.

Opening mother to reveal green means the baby MAY be in this set, but equally could be in the other set which starts Red Green - again no way of knowing which one.

That was only an example ,it may be also No. 5 if araver had put the baby there.

[araver]

WooooooW!!very interesting attempts , I said they are in decreasing sizes,and one inside the other,and logically the big one(mother)will contain the(daugher)and the daughter will contain the (child),and only one set has an additional

(baby) inside the(child).

In your 8 combinations:

No. 1 & 2 are wrong...because there are three red colored dolls under each other.

No. 4 is also wrong...because it has both two red and two green colored dolls.

No. 8 is wrong ...because it has 3 greens successively.

So the only right answer(that was in my head)is:

3,6 and 7

and the baby is in No. 6

Thank you.....you are GREAT!!!

Dear Wolfgang, I must say that I really don't understand your thoughts on this puzzle.

Did you just agree with me that there is no solution to the puzzle?

[wolfgang]

Dear Wolfgang, I must say that I really don't understand your thoughts on this puzzle.

Did you just agree with me that there is no solution to the puzzle?

Yes, I agree,,,my bad...