Find the number of non negative solutions of the system 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 $$a + b + c + d + ..... + x + y + z = 208.$$
A.
$$^{22}{P_{10}}$$
B.
$$^{22}{P_{11}}$$
C.
$$^{22}{P_{13}}$$
D.
None of these
Answer :
$$^{22}{P_{13}}$$
Solution :
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 solutions of this equation is 12.
Similarly for third equation $$a + b + c + d + e + f = 33$$ or $$e + f = 12$$ or number of solutions 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 $$11 \times 12 \times 13 \times ..... \times 21 \times 22 \times 23 = \frac{{23!}}{{10!}} = {\,^{23}}{P_{13}}.$$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is