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Permutations and Combinations question

2022 · 25 Jun · Shift 1 · Q43
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Permutations and Combinations question

2022 · 25 Jun · Shift 1 · Q43

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
Let A be a 3 ×\times× 3 matrix having entries from the set {−-− 1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 414

  1. Let the numbers of entries equal to 1,0,−11,0,-11,0,−1 in the 3×33\times 33×3 matrix be x,y,zx,y,zx,y,z respectively.

Since there are 999 entries total, x+y+z=9.x+y+z=9.x+y+z=9.

Also, the sum of all entries is 555, so x−z=5.x-z=5.x−z=5.

We must count all nonnegative integer solutions (x,y,z)(x,y,z)(x,y,z) satisfying these equations.

  1. Solve the system.

From x−z=5  ⟹  x=z+5.x-z=5 \implies x=z+5.x−z=5⟹x=z+5. Substitute into x+y+z=9:$ (z+5)+y+z=9 \implies y+2z=4.ThusThusThusy=4-2z.$$ Since y≥0y\ge 0y≥0, possible values of zzz are:

  • z=0⇒y=4, x=5z=0 \Rightarrow y=4,\ x=5z=0⇒y=4, x=5
  • z=1⇒y=2, x=6z=1 \Rightarrow y=2,\ x=6z=1⇒y=2, x=6
  • z=2⇒y=0, x=7z=2 \Rightarrow y=0,\ x=7z=2⇒y=0, x=7

So there are three cases.

  1. Count matrices in each case.

For fixed (x,y,z)(x,y,z)(x,y,z), the number of matrices is the number of ways to place:

  • xxx ones,
  • yyy zeros,
  • zzz minus ones among 999 positions: 9!x!y!z!.\frac{9!}{x!y!z!}.x!y!z!9!​.

Now evaluate each case.

Case 1: (x,y,z)=(5,4,0)(x,y,z)=(5,4,0)(x,y,z)=(5,4,0) 9!5!4!0!=(95)=126.\frac{9!}{5!4!0!}=\binom{9}{5}=126.5!4!0!9!​=(59​)=126.

Case 2: (x,y,z)=(6,2,1)(x,y,z)=(6,2,1)(x,y,z)=(6,2,1) 9!6!2!1!=362880720⋅2=252.\frac{9!}{6!2!1!}=\frac{362880}{720\cdot 2}=252.6!2!1!9!​=720⋅2362880​=252.

Case 3: (x,y,z)=(7,0,2)(x,y,z)=(7,0,2)(x,y,z)=(7,0,2) 9!7!0!2!=(92)=36.\frac{9!}{7!0!2!}=\binom{9}{2}=36.7!0!2!9!​=(29​)=36.

  1. Add all possibilities: 126+252+36=414.126+252+36=414.126+252+36=414.

Therefore, the number of such matrices is 414.\boxed{414}.414​.

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