Get Answers to all your Questions

header-bg qa

If \mathrm{ n C r: n C r+1=1: 4} and \mathrm{ n C r+1: n C r+2=4: 5}, determine the values of n and r.

Option: 1

25,000

 


Option: 2

60,000

 


Option: 3

10,000

 


Option: 4

40,000


Answers (1)

best_answer

If \mathrm{n C r: n C r+1=1: 4} and , determine the values of n and r.

Let's solve the given equations to find the values of n and r :

From the first ratio, \mathrm{n C r: n C r+1=1: 4}, we can write:

\mathrm{ n C r+1=4 * n C r }

From the second ratio, \mathrm{n C r+1: n C r+2=4: 5}, we can write:

\mathrm{ n C r+2=5 * n C r+1 }

Now, we can solve these equations to find the values of n and r.

Substituting the value of \mathrm{n C r+1} from the first equation into the second equation, we have:

\mathrm{ n C r+2=5 *(4 * n C r) }

Simplifying, we get:

\mathrm{ n C r+2=20 * n C r }

Dividing both sides by \mathrm{n C r}, we have:

\mathrm{ n C r+2 / n C r=20 }
Using the property of binomial coefficients, we have:

\mathrm{ (n+1) / r=20 }
Simplifying further, we get:

\mathrm{ n+1=20 r }

From the first equation, \mathrm{n C r+1=4 * n C r}, we have:

\mathrm{ n C r+1=4 * n C r }

Using the property of binomial coefficients, we have:

\mathrm{ (n+1) /(r+1)=4 }

Substituting\mathrm{ n+1=20 r} into the above equation, we get:

\mathrm{ 20 r /(r+1)=4 }
Cross-multiplying, we have:

\mathrm{ 20 r=4 r+4 }

Subtracting 4 r from both sides, we have:

\mathrm{ 16 r=4 }

Dividing both sides by 16 , we get:

\mathrm{ r=1 / 4 }

Substituting the value of r into \mathrm{n+1=20 r}, we have:

\mathrm{ n+1=20 *(1 / 4) }
Simplifying, we get:

\mathrm{ n+1=5 }

Subtracting 1 from both sides, we have:

\mathrm{n=4} Therefore, the values of \mathrm{n} and \mathrm{r} that satisfy the given ratios are:

\mathrm{ n=4 \text { and } r=1 / 4 }
Hence, the solution for the values of n and r in the given equations is \mathrm{n=4 \, \, and \, \, r=1 / 4.}

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE