Question

A teacher is showing a magic trick to his students. He places 12 coins on a table , 5 of which are heads up and 7 are tails up. He then places a blindfold over his eyes and shuffles the coins, keeping the faces up. Next, he separates the coins into two piles of 5 and 7 respectively. He flips over all the coins in the smaller pile. Show that both piles now have the same number of heads up
[Help me I am Suffering]

Answer 1

Step-by-step explanation:

Initially

5 coins = heads up

7 coins = tails up

1) After shuffling, Consider: i) New Pile of 5

Let m be the number of heads up in the new pile of 5. The number of tails up in this pile would be (5-m).

ii) New Pile of 7

Since in total there are only 5 heads up, thus in this pile there must be (5-m) heads up exactly and 7 - (5-m) = (m + 2) tails up.... [Fact 1]

2) After he flips over all the coins in the smaller "5 pile", the "opposite" happens namely

We will have:

m will be the number of "tails up" in this 'flipped' pile of 5. The number of "heads up" in this 'flipped' pile would be (5-m).

... [Fact 2]

Finally, [Fact 1] and [Fact 2] both state that the number of "heads up" in these two piles are now 'equal', namely to (5m).

Answer 2

Given:

Coins placed on the table$=12$

Number of heads$=5$

Number of tails$=7$

To show: Both piles now have the same number of heads up.

Solution:

Know that from the question, Initially on the table there were,

5 coins $=$ heads up

7 coins$=$ tails up

Observe that, After shuffling there are two piles of 5 and 7 coins respectively.

Consider:

i) New Pile of 5

Assume that m is the number of heads up in the new pile of 5, then The number of tails up in this pile would be (5 - m).

ii) New Pile of 7

Understand that in total there are only 5 heads up, therefore in this pile there must be $(5 - m)$ heads up exactly and $7 - (5 - m) = (m + 2)$ tails up.    and  After he flips over all the coins in the smaller "5 pile", the "opposite" happens namely.

Therefore, m will be the number of "tails up" in this 'flipped' pile of 5 and The number of "heads up" in this 'flipped' pile would be $(5 - m).$  

Observe that, the number of "heads up" in these two piles are now 'equal', namely to $(5 - m).$

Hence, it is shown that after flipping both piles have the same number of heads up.