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Sagot :
C(n,4) = 126
To solve, we need to use combination formula:
[tex]nCr = \frac{n!}{(n-r)! r!} [/tex]
Substitute the given, the solve;
[tex] \frac{n!}{(n-4)!4!}=126 [/tex]
[tex] \frac{n!}{(n-4)! 24} =126 [/tex]
[tex] \frac{n!}{(n-4)!}=126(24) [/tex]
[tex] \frac{n!}{(n-4)!} = 3024 [/tex]
[tex] \frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!} = 9(8)(7)(6) [/tex]
Cancel (n-4)! since it is common to both numerator and denominator.
So,
[tex]n(n-1)(n-2)(n-3) = 9(8)(7)(6)[/tex]
Since the left side has the same pattern with the right side, then it follows that n = 9.
For more learning about combination: https://brainly.ph/question/504738
To solve, we need to use combination formula:
[tex]nCr = \frac{n!}{(n-r)! r!} [/tex]
Substitute the given, the solve;
[tex] \frac{n!}{(n-4)!4!}=126 [/tex]
[tex] \frac{n!}{(n-4)! 24} =126 [/tex]
[tex] \frac{n!}{(n-4)!}=126(24) [/tex]
[tex] \frac{n!}{(n-4)!} = 3024 [/tex]
[tex] \frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!} = 9(8)(7)(6) [/tex]
Cancel (n-4)! since it is common to both numerator and denominator.
So,
[tex]n(n-1)(n-2)(n-3) = 9(8)(7)(6)[/tex]
Since the left side has the same pattern with the right side, then it follows that n = 9.
For more learning about combination: https://brainly.ph/question/504738
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