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QJEEAdv17Ma300e: JEE Advanced 2017 Mathematics - Class 12 JEEAdv17Ma300e Solution: Three randomly chosen nonnegative integers x, y and z are found …

Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is

Multiple Choice Options

  • A 12{1 \over {2}}
  • B 3655{36 \over {55}}
  • C 611{6 \over {11}}
  • D 511{5 \over {11}}

Step-by-Step Verified Solution

To solve this problem, we need to find the total number of possible solutions to the equation x+y+z=10x + y + z = 10 where x,y,x, y, and zz are nonnegative integers.

First, we employ the "stars and bars" method to determine the number of nonnegative integer solutions to this equation. This method states that the number of ways to partition nn identical items (our sum) into kk distinct groups (our variables) is given by:

(n+k−1k−1)\binom{n + k - 1}{k - 1}

For x+y+z=10x + y + z = 10 , we have n=10n = 10 and k=3k = 3 . Plugging into the formula, we get:

(10+3−13−1)=(122)\binom{10 + 3 - 1}{3 - 1} = \binom{12}{2}

We calculate the binomial coefficient:

(122)=12×112×1=66\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66

So, there are 66 possible solutions to the equation x+y+z=10x + y + z = 10 .

Next, we need to find the number of solutions for which zz is even. Let z=2kz = 2k where kk is a nonnegative integer. Then the equation becomes:

x+y+2k=10x + y + 2k = 10

Rearranging it, we get:

x+y=10−2kx + y = 10 - 2k

Here, 10−2k10 - 2k must be nonnegative, so kk can take values 0,1,2,3,4,50, 1, 2, 3, 4, 5 making total 6 possible values of k.

For each value of kk , x+yx + y must equal the corresponding 10−2k10 - 2k . The number of nonnegative integer solutions to x+y=mx + y = m for any nonnegative integer mm is given by:

(m+11)=m+1\binom{m + 1}{1} = m + 1

We will sum the solutions for each valid kk :

When k=0k = 0 : x+y=10x + y = 10 , the number of solutions is 1111 .

When k=1k = 1 : x+y=8x + y = 8 , the number of solutions is 99 .

When k=2k = 2 : x+y=6x + y = 6 , the number of solutions is 77 .

When k=3k = 3 : x+y=4x + y = 4 , the number of solutions is 55 .

When k=4k = 4 : x+y=2x + y = 2 , the number of solutions is 33 .

When k=5k = 5 : x+y=0x + y = 0 , the number of solutions is 11 .

Summing these, we get:

11+9+7+5+3+1=3611 + 9 + 7 + 5 + 3 + 1 = 36

Thus, there are 36 solutions where zz is even out of a total of 66 solutions. Hence, the probability that zz is even is given by:

3666=611\frac{36}{66} = \frac{6}{11}

Thus, the correct answer is:

Option C: 611\frac{6}{11}

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