Question: Find the number of non-negative solutions of the systems of equations: $ a+b=10,a+b+c+d=21,a+b+c+d+e+f=33, $ $ a+b+c+d+e+f+g+h=46 $ , and so on till $ a+b+c+d+….+x+y+z=208. $
Options:
$ ^{10}P _{22} $
B) $ ^{22}_{11}\text{P} $
C) $ ^{13}_{22}P $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Consider the equation $ a+b=10 $ number of solutions of this equation is $ ^{10+2-1}{C _{2-1}}=11. $ Next equation is $ a+b+c+d=21 $ hence $ c+d=11 $ and number of solution of this equation is 12. Similarly for third equation $ a+b+c+d+e+f=33 $ or $ e+f=12 $ or number of solution is 13. Similarly for last equation $ a+b+c+d+…+x+y+z=208, $ or $ y+z=22 $ or number of solution is 23. Required number of ways is $ 11\times 12\times 13\times …\times 21\times 22\times 23=\frac{23!}{10!} $
BETA
