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21 Sep 2019, 16:10
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
62% (02:49) correct
38%
(02:38)
wrong
based on 1977
sessions
History
Date
Time
Result
Not Attempted Yet
If n > 4, what is the value of the integer n ?
(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)
(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)
DS47661.01
(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)
(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)
DS47661.01
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19 Oct 2019, 01:05
Kudos
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If n > 4, what is the value of the integer n ?
(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)
Two ways..
(a) Combination formula
\(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\).....\(\frac{n!}{3!(n - 3)!} = \frac{n!}{4!(n - 4)!}.....nC3=nC4\)
Thus \(n=3+4=7\)..SUFF
(b) Arithmetic way
\(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\).......\(\frac{n(n-1)(n-2)(n-3)!}{(n - 3)!} = \frac{3!n(n-1)(n-2)(n-3)(n-4)!}{4!(n - 4)!}\)....\(4!*n(n-1)(n-2)=3!*(n)(n-1)(n-2)(n-3).....4=n-3...n=7..SUFF\)
(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)
Two ways..
(a) Combination formula
\(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\).....\(.....nC3+nC4=(n+1)C4\)
Thus n could be anything as it is true for all.
(b) Arithmetic way
\(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\) ......
Take out n!/3!(n-3)!
\(\frac{n!}{3!(n-3)!}(1+\frac{n-3}{4} )=\frac{n!}{3!(n-3)!}(\frac{n+1}{4} ).....\frac{n+1}{4}=\frac{n+1}{4}\)....Always true..
This also tells us why nC3+nC4=(n+1)C4
A
(1) \(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\)
Two ways..
(a) Combination formula
\(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\).....\(\frac{n!}{3!(n - 3)!} = \frac{n!}{4!(n - 4)!}.....nC3=nC4\)
Thus \(n=3+4=7\)..SUFF
(b) Arithmetic way
\(\frac{n!}{(n - 3)!} = \frac{3!n!}{4!(n - 4)!}\).......\(\frac{n(n-1)(n-2)(n-3)!}{(n - 3)!} = \frac{3!n(n-1)(n-2)(n-3)(n-4)!}{4!(n - 4)!}\)....\(4!*n(n-1)(n-2)=3!*(n)(n-1)(n-2)(n-3).....4=n-3...n=7..SUFF\)
(2) \(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\)
Two ways..
(a) Combination formula
\(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\).....\(.....nC3+nC4=(n+1)C4\)
Thus n could be anything as it is true for all.
(b) Arithmetic way
\(\frac{n!}{3!(n - 3)!} + \frac{n!}{4!(n - 4)!} = \frac{(n + 1)!}{4!(n - 3)!}\) ......
Take out n!/3!(n-3)!
\(\frac{n!}{3!(n-3)!}(1+\frac{n-3}{4} )=\frac{n!}{3!(n-3)!}(\frac{n+1}{4} ).....\frac{n+1}{4}=\frac{n+1}{4}\)....Always true..
This also tells us why nC3+nC4=(n+1)C4
A
03 Feb 2021, 03:03
Kudos
Bookmarks
Don’t get flummoxed by complicated equations and factorials. Try to break down factorials to cancel a few of them from both sides.
Check the attachment. Answer is A.
Hope this helps.
Check the attachment. Answer is A.
Hope this helps.
Attachments
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