test series(combinatorics)

Question

Q.1) The number of integral solution of equation X1+X2+X3+X4=20 where X1>=3,X2>=1,X3>=0,X4>=5  is...................

can anyone explain the approach to solve this question???

Comments

may this is useful https://gateoverflow.in/123727/isi-2004-miii

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thanx

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@BASANT KUMAR

what is the answer ?

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The equation becomes-

X1+X2+X3+X4=11

so using stars and bars,

$\binom{11+3}{3}=\binom{14}{3}=364$

correct?

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Yes $364$

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@Swapnil Naik

@Lakshman Patel RJIT

how it is X1 +X2 + X3 +X4=11

plz explain?

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Originally x1+x2+x3+x4= 20, this means we need to select values of x1,x2,x3,x4 which will sum up to 20. This includes 0,0,0,20 then 0,0,1,19, ...... and all possible combinations. But in given case there are restrictions on x i.e. based on their minimum value. x1,x2,x3,x4 can have minimum values 3,1,0,5 respectively. Hence we can't use combination like this 0,0,0,20. In any combination we must have at least 3x1,1x2,0x3,5x4.

 If you write  x1+x2+x3+x4=11, the combinations you can have 0,0,0,11 and so on... Now add minimum values 0+3,0+1,0+0,11+5 = 3,1,0,16 which is a valid combination and if you know 9 items are minimum then for the rest 11 we need to find combinations.

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Good Brother,i understand the concept.

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@Swapnil Naik

brother still not getting this

 if you know 9 items are minimum then for the rest 11 we need to find combinations.

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@himgta

see my answer.it might be useful

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yes 364 is correct answer.

Answer 1

The approach would be to subtract the offset part from the total sum and formulate a new problem as below :

y1 + y2 + y3 + y4 = 11, and y1, y2, y3, y4 >= 0.

Now try to solve this question by finding out the number of ways you can put 3 partitions between 11 similar items. This will give you ways to divide 11 items in 4 parts, where each part will correspond to y1, y2, y3 and y4.

Note that in this particular problem there can be only one non-zero y too. That will happen when all 3 partitions are put together at the end or beginning of these 11 items.

Answer 2

Given that:$X_{1}+X_{2}+X_{3}+X_{4}=20$   where $X_{1}\geq3,X_{2}\geq1,X_{3}\geq0,X_{4}\geq5 $

we can write like this :

$X_{1}+X_{2}+X_{3}+X_{4}=20$-------->$(1)$   where $X_{1}-3\geq0,X_{2}-1\geq0,X_{3}\geq0,X_{4}-5\geq0$

Let $X_{1}-3=Y_{1},X_{2}-1=Y_{2},X_{3}=Y_{3},X_{4}-5=Y_{4}$

      $X_{1}=Y_{1}+3,X_{2}=Y_{2}+1,X_{3}=Y_{3},X_{4}=Y_{4}+5$

Put the value in the equation $(1),$

          $Y_{1}+3+Y_{2}+1+Y_{3}+Y_{4}+5=20$  where $Y_{1}\geq0,Y_{2}\geq0,Y_{3}\geq0,Y_{4}\geq0$

    $\Rightarrow Y_{1}+Y_{2}+Y_{3}+Y_{4}+9=20$   where $Y_{1}\geq0,Y_{2}\geq0,Y_{3}\geq0,Y_{4}\geq0$

  $\Rightarrow Y_{1}+Y_{2}+Y_{3}+Y_{4}=11$------>$(2)$   where $Y_{1}\geq0,Y_{2}\geq0,Y_{3}\geq0,Y_{4}\geq0$

The Number of Integral Solution of Equation $=\binom{n+r-1}{r}$ where $n=$Number of Variables and $r=$Sum of the variables.

Here $n=4$ and $r=11$

                                $\Rightarrow\binom{4+11-1}{11}$

                                $\Rightarrow\binom{14}{11}$

                               $\Rightarrow\binom{14}{3}=364$

Comments

Thanks!