Question

invent the value of n if nc5=42nc3 where n less than 4​

Answer 1

SOLUTION

CORRECT QUESTION

Invent the value of n if

$\sf{ {}^{n} P_5 = 42 \: {}^{n}P_3 \: \: \: \: \: where \: n > 4 }$

EVALUATION

Here it is given that

$\sf{ {}^{n} P_5 = 42 \: {}^{n}P_3 \: }$

$\displaystyle \sf{ \implies \: \frac{n!}{(n - 5)!} = 42 \times \frac{n!}{(n - 3)!}}$

$\displaystyle \sf{ \implies \: \frac{(n - 3)!}{(n - 5)!} = 42}$

$\displaystyle \sf{ \implies \: \frac{(n - 3) \times (n - 4) \times (n - 5)!}{(n - 5)!} = 42}$

$\displaystyle \sf{ \implies \:(n - 3) \times (n - 4) = 42}$

$\displaystyle \sf{ \implies \: {n }^{2} - 7n + 12= 42}$

$\displaystyle \sf{ \implies \: {n }^{2} - 7n - 30 = 0}$

$\displaystyle \sf{ \implies \: {n }^{2} - (10 - 3)n - 30 = 0}$

$\displaystyle \sf{ \implies \: {n }^{2} - 10 n + 3n- 30 = 0}$

$\displaystyle \sf{ \implies \: n(n - 10) + 3(n- 10 )= 0}$

$\displaystyle \sf{ \implies \: (n - 10) (n + 3)= 0}$

$\displaystyle \sf{ \implies \: either \: \: (n - 10) = 0 \: \: or \: \: (n + 3)= 0}$

Now

$\displaystyle \sf{ \: (n - 10) = 0 \: \: \: gives \: \: \: n = 10}$

$\displaystyle \sf{ \: (n + 3) = 0 \: \: \: gives \: \: \: n = - 3}$

Since n can not be negative

$\sf{ \therefore \: \: n \: \ne \: - 3}$

$\sf{ \therefore \: \: n \: = \: 10}$

FINAL ANSWER

The required value of n is 10

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