GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Number of distinct four-digit numbers that can be formed VS 560
[#permalink]
24 Aug 2021, 04:06
24 Aug 2021, 04:06
1
2
22
Bookmarks
00:00
Question Stats:
55% (01:43) correct
44% (01:48) wrong based on 49 sessions
Hide Show timer Statistics
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
[#permalink]
31 Aug 2021, 21:57
31 Aug 2021, 21:57
3
2
Bookmarks
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?
