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Byju's Answer

Standard IX

Mathematics

Ratio and Proportion

Eight cards b...

Question

Eight cards bearing number $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ are well shuffled. Then in how many cases the top $2$ cards will form a pair of twin prime equals

720

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Solution

The correct option is C $720$
Total no. of cards $8$
There are few cares by which the top card has $2$ twin prices
$1) . (5, 7)$ is a twin prime.
$2) . (3, 5)$ is a twin prime.
The equations are of the form $(6 n - 1, 6 n + 1)$
$\Rightarrow$ No. of ways to select $= 20 \times 6 \times 6 = 720.$
Hence, the answer is $720.$

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Eight cards bearing number 1, 2, 3, 4, 5, 6, 7, 8 are well shuffled. Then in how many cases the top 2 cards will form a pair of twin prime equals

The correct option is C 720Total no. of cards 8There are few cares by which the top card has 2 twin prices 1).(5,7) is a twin prime.2).(3,5) is a twin prime.The equations are of the form (6n−1,6n+1)⇒ No. of ways to select =20×6×6=720.Hence, the answer is 720.

Eight cards bearing number $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ are well shuffled. Then in how many cases the top $2$ cards will form a pair of twin prime equals