Explanation:Case I: All numbers are different i.e. using 1, 2, 3, 4, 5, and 6Number of ways of selecting 4 numbers out of 6 = \(^6C_4 = 15\)
Number of ways in which we can arrange these = \(4! = 24\)
Total number of ways = \((15)(24) = 360\)
Case II: two 5's and two different numbers i.e. using 1, 2, 3, 4, 6, 5, and 5Number of ways of selecting two 5's numbers out of 5 = \(^5C_2 = 10\)
Number of ways in which we can arrange these = \(\frac{4!}{2!} = 12\)
Total number of ways = \((10)(12) = 120\)
Case III: two 6's and two different numbers i.e. using 1, 2, 3, 4, 5, 6 and 6Number of ways of selecting two 6's numbers out of 5 = \(^5C_2 = 10\)
Number of ways in which we can arrange these = \(\frac{4!}{2!} = 12\)
Total number of ways = \((10)(12) = 120\)
Case IV: two 5's and two 6's i.e. 5, 5, 6 and 6Number of ways in which we can arrange these = \(\frac{4!}{(2!)(2!)} = 6\)
Col. A: 360 + 120 + 120 + 6 = 606
Col. B: 560
Hence, option A
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/