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Number of distinct four-digit numbers that can be formed VS 560
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24 Aug 2021, 04:06
24 Aug 2021, 04:06
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00:00
Question Stats:
50% (01:48) correct
49% (01:47) wrong based on 85 sessions
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Quantity A |
Quantity B |
Number of distinct four-digit numbers that can be formed by using the digits - 1, 2, 3, 4, 5, 5, 6, and 6 |
560 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: Number of distinct four-digit numbers that can be formed VS 560
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31 Aug 2021, 21:57
31 Aug 2021, 21:57
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Explanation:
Case I: All numbers are different i.e. using 1, 2, 3, 4, 5, and 6
Number 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 5
Number 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 6
Number 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 6
Number 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
Case I: All numbers are different i.e. using 1, 2, 3, 4, 5, and 6
Number 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 5
Number 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 6
Number 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 6
Number 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
General Discussion
Re: Number of distinct four-digit numbers that can be formed VS 560
[#permalink]
24 Nov 2023, 01:44
24 Nov 2023, 01:44
Why isn't the solution 6X6x6x6?
