Class 12 Permutation and Combination Exercise 1.3 Solutions | Basic Mathematics Sukunda Pustak Bhawan
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If C(20, r+5) = C (20, 2r - 7), find C(15, r).
Atith Adhikari
·
2 years ago
Solution
If $\rm C (n, r) = C(n, r \prime)$, then either $\rm r = r \prime$ or $\rm r + r \prime = n$.
For the equation, $\rm C(20, r + 5) = C (20, 2r - 7)$, we have $\rm n = 20$, $\rm r = r + 5$ and $\rm r \prime = 2r - 7$.
If $\rm r = r \prime$
$\rm r + 5 = 2 r - 7$
$\rm or, r + 5 + 7 = 2r - 7 + 7$
$\rm or, r + 12 = 2r$
$\rm or, r -r + 12 = 2r -r $
$\rm or, 12 = r$
$\rm \therefore r = 12$
Putting $\rm r = 12$ in the given equation $\rm C(20, r + 5) = C(20, 2r - 7)$, the equation holds. Hence, the required value of r is 12.
$\rm \therefore C(15, r) = C(15, 12) = 455$
If $\rm r + r \prime = n$
$\rm (r + 5) + (2r - 7) = 20$
$\rm or, r + 2r + 5 -7 = 20$
$\rm or, 3r - 2 = 20$
$\rm or, 2r -2 +2 = 20 +2$
$\rm or, 2r = 22$
$\rm \therefore r = 11$
Putting $\rm r = 11$in the given equation $\rm C(20, r + 5) = C(20, 2r - 7)$, the equation does not hold. Hence, the required value of r cannot be 11.
If $ \rm C(n, 10) + C(n, 9) = C(20, 10) $ find $\rm n$ and $\rm C(n, 17)$
Atith Adhikari
·
2 years ago
Solution
Given
$\rm C(n,10) + C(n,9) = C(20,10)$. This equation follows Pascal's identity in Combinations which states
$$\rm C(n, r) + C(n, r-1) = C(n+1,r)$$
By comparison, we get, $\rm n = n$, $\rm r = 10$, $\rm n+1 = 20$
Solving for n in the last equation, we get,
$\rm n + 1 = 20$
$\rm \therefore n = 19$
Hence, the required value of n is 19.
Again, to find $\rm C(n,17)$, we use the formula
$$\rm C(n, r) = \frac{n!}{(n-r)! \cdot r!}$$
$\rm \therefore C(19, 17) = \frac{19!}{(19 -17)! \cdot 17!} = 171$.
Hence, the required value of $\rm C(19, 17)$ is 171.
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