Six boys A, B, C, D, E and F play a game of cards. Each has a pack of 10 cards. F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes on 1 to E. Then the number of cards possessed by D and E is equal to the number of cards possessed by

The correct option is B, C and F.

Let us break down the problem step-by-step to track the exact number of cards held by each boy after all transactions take place.

Initial State:
Each of the 6 boys (A, B, C, D, E, F) starts with 10 cards.
• A = 10
• B = 10
• C = 10
• D = 10
• E = 10
• F = 10

Transactions:

1. F borrows 2 cards from A:
• A loses 2 cards: $10-2=8$
• F gains 2 cards: $10+2=12$

2. F gives away 5 cards to C:
• F loses 5 cards: $12-5=7$
• C gains 5 cards: $10+5=15$

3. C gives 3 cards to B:
• C loses 3 cards: $15-3=12$
• B gains 3 cards: $10+3=13$

4. B gives 6 cards to D:
• B loses 6 cards: $13-6=7$
• D gains 6 cards: $10+6=16$

5. D passes on 1 card to E:
• D loses 1 card: $16-1=15$
• E gains 1 card: $10+1=11$

Final Card Counts:
• A = 8
• B = 7
• C = 12
• D = 15
• E = 11
• F = 7

Comparison:
The total number of cards possessed by D and E combined is:
$\text{D}+\text{E}=15+11=26$

Now, let us calculate the combined number of cards for the combinations given in the options:

• A, B and C: $8+7+12=27$
• B, C and F: $7+12+7=26$
• A, B and F: $8+7+7=22$
• A, C and F: $8+12+7=27$

Therefore, the number of cards possessed by D and E combined (26) is equal to the number of cards possessed by B, C and F.