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  • The number of ordered pairs (r,k) for which <img alt="6\cdot ^{35}C_{r}=(k^{2}-3)\cdot ^{36}C_{r+1}," src="http://entrancecorner.oncodecogs.com/gif.latex?6%5Ccdot%20%5E%7B35%7DC_%7Br%7D%3D%28k%5E%7B2%7D-3%29%5Ccdot%20%5E%7B36%7DC_%7Br&plus;1%7

The number of ordered pairs (r,k) for which 6\cdot ^{35}C_{r}=(k^{2}-3)\cdot ^{36}C_{r+1}, where k is an integer, is :
Option: 1 4
Option: 2 6
Option: 3 2
Option: 4 3
 

Answers (1)

best_answer

As we have learnt

 ^{n} C_{r}=\frac{n}{r} \cdot^{n-1} C_{r-1}

   

Now, 

^{36}C_{r+1} (k^{2}-3)={^{35}C_{r}}\times {6}

\frac{36}{r+1}. ^{35}C_{r} (k^{2}-3)={^{35}C_{r}}\times {6}

{k^{2}-3=\frac{r+1}{6} \Rightarrow k^{2}=3+\frac{r+1}{6}}

r should be less than or equal to 35

Hence for k to be an integer, r can be 5 and 35

For r=5 we get k = -2, 2

For r=35 we get k=-3,3

We get 4 ordered pair (5,-2), (5,2), (35,-3), (35, 3)

Correct Option (1)

Posted by

Ritika Jonwal

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