Question

How many solutions do the following equation have

x₁+x₂+x₃=11

where x₁≥1, x₂≥2, x₃≥3

(A) C(7, 11)       (B) C(11, 3)

(C) C(14, 11)     (D) C(7, 5)

Answer 1

D) is answer

Given that

$x1+x2+x3=11$ where, $x1\geqslant 1,x2\geqslant 2,x3\geqslant 3$

Let $x1=x1{}'+1, x2=x2{}'+2, x3=x3{}'+3$

Using above equation we can write $x1{}'+1+x2{}'+2+x3{}'+3=11 \Rightarrow x1{}'+x2{}'+x3{}'=5$

Now it is the case of combination with repition which is given by $\binom{n+r-1}{r}$

so Answer is $\binom{3+5-1}{5}=\binom{7}{5}=\binom{7}{2}$

Answer 2

we can solve this problem by combinition with repetition.

so we need to put 11 similar balls into  x1,x2,x3 types of boxes.

different number of boxes are 3=n

x₁≥1, x₂≥2, x₃≥3 now allocate 1,2,3 from 11 directly . so remaining  ball = 11-(1+2+3) = 5 =k

(n+k-1)C_k  = (3+5-1)C5 =7C5