Let n_1\, \lt n_2\, \lt \,n_3\, \lt \,n_4\, \lt n_5 be positive integers such that n_1\, + n_2\, + \,n_3\, + \,n_4\, + n_5 = 20. Then the number of such destinct arrangements \,(n_1\,,\,n_2,\,\,n_3,\,\,n_4\,,n_5) is

Permutations and Combinations PYQ — JEE Advanced 2014 · Shift 1 · Q21 with Solution

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Correct answer: 7

We need the number of 5-tuples of positive integers $n_1<n_2<n_3<n_4<n_5$ such that $n_1+n_2+n_3+n_4+n_5=20.$

Since the inequalities are strict, this is equivalent to choosing 5 distinct positive integers whose sum is 20.

Convert strict inequalities into weak inequalities.

Define

\quad m_2=n_2-1, \quad m_3=n_3-2, \quad m_4=n_4-3, \quad m_5=n_5-4.$$ Then from $$n_1<n_2<n_3<n_4<n_5,$$ we get $$m_1\ge 1, \quad m_2\ge 1, \quad m_3\ge 1, \quad m_4\ge 1, \quad m_5\ge 1,$$ and also $$m_1\le m_2\le m_3\le m_4\le m_5.$$ Their sum is $$m_1+m_2+m_3+m_4+m_5=(n_1+n_2+n_3+n_4+n_5)-(0+1+2+3+4)=20-10=10.$$ So we need the number of partitions of 10 into 5 positive parts. 3. Since all 5 parts are positive, subtract 1 from each: $$x_i=m_i-1 \ge 0.$$ Then $$x_1+x_2+x_3+x_4+x_5=10-5=5,$$ with $$x_1\le x_2\le x_3\le x_4\le x_5.$$ Thus we need the number of partitions of 5 into at most 5 parts, which is just the number of partitions of 5. 4. List all partitions of 5: $$5$$ $$4+1$$ $$3+2$$ $$3+1+1$$ $$2+2+1$$ $$2+1+1+1$$ $$1+1+1+1+1$$ There are $7$ such partitions. Hence the required number of strictly increasing 5-tuples is $$\boxed{7}.$$ 5. Verification by direct construction: Each partition of 5 gives a nondecreasing 5-tuple $(x_1,x_2,x_3,x_4,x_5)$, then $$m_i=x_i+1,$$ and finally $$n_i=m_i+(i-1).$$ For example: - $5 \to (0,0,0,0,5) \to (1,1,1,1,6) \to (1,2,3,4,10)$ - $4+1 \to (0,0,0,1,4) \to (1,1,1,2,5) \to (1,2,3,5,9)$ - $3+2 \to (0,0,0,2,3) \to (1,1,1,3,4) \to (1,2,3,6,8)$ - $3+1+1 \to (0,0,1,1,3) \to (1,1,2,2,4) \to (1,2,4,5,8)$ - $2+2+1 \to (0,0,1,2,2) \to (1,1,2,3,3) \to (1,2,4,6,7)$ - $2+1+1+1 \to (0,1,1,1,2) \to (1,2,2,2,3) \to (1,3,4,5,7)$ - $1+1+1+1+1 \to (1,1,1,1,1) \to (2,2,2,2,2) \to (2,3,4,5,6)$ Exactly 7 tuples arise. Therefore, the answer is correct.

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View written solutionFree

Correct answer: 7

We need the number of 5-tuples of positive integers $n_1<n_2<n_3<n_4<n_5$ such that $n_1+n_2+n_3+n_4+n_5=20.$

Since the inequalities are strict, this is equivalent to choosing 5 distinct positive integers whose sum is 20.

Convert strict inequalities into weak inequalities.

Define

\quad m_2=n_2-1, \quad m_3=n_3-2, \quad m_4=n_4-3, \quad m_5=n_5-4.$$ Then from $$n_1<n_2<n_3<n_4<n_5,$$ we get $$m_1\ge 1, \quad m_2\ge 1, \quad m_3\ge 1, \quad m_4\ge 1, \quad m_5\ge 1,$$ and also $$m_1\le m_2\le m_3\le m_4\le m_5.$$ Their sum is $$m_1+m_2+m_3+m_4+m_5=(n_1+n_2+n_3+n_4+n_5)-(0+1+2+3+4)=20-10=10.$$ So we need the number of partitions of 10 into 5 positive parts. 3. Since all 5 parts are positive, subtract 1 from each: $$x_i=m_i-1 \ge 0.$$ Then $$x_1+x_2+x_3+x_4+x_5=10-5=5,$$ with $$x_1\le x_2\le x_3\le x_4\le x_5.$$ Thus we need the number of partitions of 5 into at most 5 parts, which is just the number of partitions of 5. 4. List all partitions of 5: $$5$$ $$4+1$$ $$3+2$$ $$3+1+1$$ $$2+2+1$$ $$2+1+1+1$$ $$1+1+1+1+1$$ There are $7$ such partitions. Hence the required number of strictly increasing 5-tuples is $$\boxed{7}.$$ 5. Verification by direct construction: Each partition of 5 gives a nondecreasing 5-tuple $(x_1,x_2,x_3,x_4,x_5)$, then $$m_i=x_i+1,$$ and finally $$n_i=m_i+(i-1).$$ For example: - $5 \to (0,0,0,0,5) \to (1,1,1,1,6) \to (1,2,3,4,10)$ - $4+1 \to (0,0,0,1,4) \to (1,1,1,2,5) \to (1,2,3,5,9)$ - $3+2 \to (0,0,0,2,3) \to (1,1,1,3,4) \to (1,2,3,6,8)$ - $3+1+1 \to (0,0,1,1,3) \to (1,1,2,2,4) \to (1,2,4,5,8)$ - $2+2+1 \to (0,0,1,2,2) \to (1,1,2,3,3) \to (1,2,4,6,7)$ - $2+1+1+1 \to (0,1,1,1,2) \to (1,2,2,2,3) \to (1,3,4,5,7)$ - $1+1+1+1+1 \to (1,1,1,1,1) \to (2,2,2,2,2) \to (2,3,4,5,6)$ Exactly 7 tuples arise. Therefore, the answer is correct.

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