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Find n

[tex]\\ \rm\Rrightarrow {}^2nC_3:{}^nC_3=12:1[/tex]

Note:- It's C represents combination not Carbon.


Kindly don't add your irrelevant or unexplained answers if you are not sure .​


Sagot :

The value of [tex]n[/tex] is [tex]5[/tex].

Step-by-step Explanation:

[tex] \frac{(2n - 1)(2n - 2)}{3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{n(n - 1)(n - 2)} = \frac{12}{1}[/tex]

[tex] = > 4(2n - 1) = 12(n - 2)[/tex]

[tex] = > 2n - 1 = 3n - 6[/tex]

[tex] = > 2n - 3n = - 6 + 1[/tex]

[tex] = > - n = - 5[/tex]

Cancelling common negative sign,

[tex] = > n = 5[/tex]

[tex] \\ \purple{\maltese}\large\underline{\underline{\sf\:\: Solution :}}\\ \\ [/tex]

here we have ,

[tex]\\ \sf \implies \: {}^{2n} \: C \: _3: \: {}^ {n} \: C _3 \: = \: 12 \: : \: 1 \\ [/tex]

[tex]\\ \sf \implies \: \frac{2 \: n \: ! }{3 \: ! \: (2n - 3) \: } \div \frac{ \: n \: ! }{3 \: ! \: (n - 3) \: } = \frac{12}{1} \\ [/tex]

[tex]\\ \sf \implies \: \frac{2 \: n \: ! }{3 \: ! \: (2n - 3) \: } \times \frac{3 \: ! \: (n - 3) \: }{\: n \: !} = \frac{12}{1} \\ [/tex]

[tex]\\ \sf \implies \: \frac{2 \: n \: (2n - 1) \: \: (2n - 2) \: \: (2n - 3) \: ! }{3 \: ! \: (2n - 3) \: !} \div \frac{3 \: ! \: (n - 3) \: ! }{2 \: n \: (2n - 1) \: \: (2n - 2) \: \: (2n - 3) \: ! } = \frac{12}{1} \\ [/tex]

[tex]\\ \sf \implies \: \frac{2 \: n \: (2n - 1) \: \: (2n - 2) \: \: \: }{n\: (n - 1)(n - 2) \: } = \frac{12}{1} \\ [/tex]

[tex]\\ \sf \implies \: \frac{2 \times 2 \: (n - 1) \: \: (2n - 1) \: \: \: }{\: (n - 1)(n - 2) \: } = \frac{12}{1} \\ [/tex]

[tex]\\ \sf \implies \: \frac{4 \: \: (2n - 1) \: \: \: }{\: (n - 2) \: } = \frac{12}{1} \\ [/tex]

[tex]\\ \sf \implies \: \frac{ \: \: (2n - 1) \: \: \: }{\: (n - 2) \: } = \frac{12}{4} \\ [/tex]

[tex]\\ \sf \implies \: \frac{ \: \: (2n - 1) \: \: \: }{\: (n - 2) \: } = \cancel\frac{12}{4} \\ [/tex]

[tex]\\ \sf \implies \: \frac{ \: \: (2n - 1) \: \: \: }{\: (n - 2) \: } = \: \frac{ \: 3 \: }{ \: 1 \: } \\ [/tex]

[tex]\\ \sf \implies \: 2n - 1 = 3 \: n - 6 \\ [/tex]

[tex]\\ \sf \implies \: 3n - 2n = 6 -3 \\ [/tex]

[tex]\\ \sf \implies \: n= 5 \\ [/tex]

[tex]\\\large\implies\sf \underline{ \boxed{\sf n=5}} \\ [/tex]

Henceforth, the value of n is 5 .