The Bird Watching Program : Puzzle

Abel, Mabel, and Caleb went bird watching. Each of them saw one bird that none of the others did. Each pair saw one bird that the third did not. And one bird was seen by all three. Of the birds Abel saw, two were yellow. Of the birds Mabel saw, three were yellow. Of the birds Caleb saw, four were yellow. 

How many yellow birds were seen in all? How many non-yellow birds were seen in all?

Click here for SOLUTION! 

The Bird Watching Program Puzzle : Solution

What was the puzzle?

Let's find the total count of birds including yellow and non-yellow.

1. Each of them saw one bird that none of the others did. Individuals - Abel, Mabel, and Caleb. So, Birds = 3.

2. Each pair saw one bird that the third did not. Pairs - Abel-Mabel, Abel-Caleb, and Mabel-Caleb. So, Birds = 3.

3. And one bird was seen by all three. Abel-Mabel-Caleb. Birds = 1.

Therefore, total of 3 + 3 + 1 = 7 birds were seen.

It's clear that, everyone saw 4 birds viz. one individually, two when in pairs and 1 with all along. 

Since Caleb saw 4 yellow birds, Caleb must not had saw any of non-yellow birds.

The 4 yellow birds that Caleb saw must be as below.

1. The one Caleb saw individually. 

2. The one that Caleb saw in pair with Abel-Caleb.

3. The one that Caleb saw in pair with Mabel-Caleb. 

4. The one that Caleb saw with all.

Now, the two yellow birds that Abel saw must be the one that is pointed by (2) above and other pointed by (4) above.

That's make sure that the pair Abel-Mabel saw a non-yellow bird. 

Since Mabel saw 3 yellow birds, 2 of them are as pointed by (3) and (4) above, the third one must be the yellow bird that she saw alone. So far, we discovered there must be 5 yellow birds (4 seen by Caleb and one seen by Mabel individually) at least.

So, far we know Abel saw a yellow bird as in pair Abel-Caleb, one more yellow bird with all and a non-yellow bird as in pair Abel-Mabel. Hence, the fourth bird that Abel saw must be a non-yellow bird since as per data Abel saw only 2 yellow birds. 

Aliens Meeting on the Earth

100 aliens attended an intergalactic meeting on earth.

73 had two heads,
28 had three eyes,
21 had four arms,
12 had two heads and three eyes,
9 had three eyes and four arms,
8 had two heads and four arms,
3 had all three unusual features.

How many aliens had none of these unusual features?

These are Aliens not having unusual features! 

Normal Aliens in the Meeting on the Earth

Read given data first!

Let's simplify the process with Venn diagram like below.

1. We know, 3 had all three unusual features.

2. 12 had two heads and three eyes, 9 had three eyes and four arms, 8 had two heads and four arms. Here, 3 having all unusual features too need to be counted in each of Aliens' counts above. 

So, Aliens having 2 unusual features of 2 heads & 3 eyes = 12 - 3 = 9.

Aliens having 2 unusual features of 3 eyes & 4 arms = 9 - 3 = 6.

Aliens having 2 unusual features of 2 heads & 4 arms = 8 - 3 = 5.

3. We know, 73 had two heads, 28 had three eyes, 21 had four arms.

Therefore, the number of Aliens having only 2 heads as an only unusual feature is 73 - (9 + 3 + 5) = 56.  

The number of Aliens having 3 eyes as an only unusual feature is 
28 - (9 + 3 + 6) = 10.

The number of Aliens having 4 arms as an only unusual feature is 
21 - (5 + 3 + 6) = 7.

4. So, we have 56 + 9 + 10 + 3 + 5 + 6 + 7 = 96 Aliens having at least one unusual feature. 

Therefore, the number of Aliens not having any of unusual features is 
100 - 96 = 4.