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-distinct-arrangements-n-1-n-2-n-3-n-4-n-5-is-

Question Number 21933 by Tinkutara last updated on 07/Oct/17

Let n_{1} < n_{2} < n_{3} < n_{4} < n_{5} be positive

integers such that n_{1} + n_{2} + n_{3} + n_{4} +

n_{5} = 20 . Then the number of such

distinct arrangements (n_{1}, n_{2}, n_{3}, n_{4}, n_{5})

is

Commented by mrW1 last updated on 08/Oct/17

7 ?

1 2 3 4 10

1 2 3 5 9

1 2 3 6 8

1 2 4 5 8

1 2 4 6 7

1 2 4 5 6 \Rightarrow should be 1 3 4 5 7

2 3 4 5 6

Commented by Rasheed.Sindhi last updated on 08/Oct/17

1 + 3 + 4 + 5 + 7

Commented by Rasheed.Sindhi last updated on 08/Oct/17

line number last but one

1 2 4 5 6

1 + 2 + 4 + 5 + 6 = 18 ?

If this line is replaced by

1 3 4 5 7

the answer is 7 yet

Commented by Tinkutara last updated on 08/Oct/17

mrW 1, why you {don}^{'} t take 1 + 3 + 4 + 5 + 7 ?

But your answer 7 is correct .

Commented by mrW1 last updated on 08/Oct/17

Thanks to you both! I made a mistake .

1 3 4 5 7 is correct, not 1 2 4 5 6 .

Commented by mrW1 last updated on 09/Oct/17

Certainly we {don}^{'} t need to list all the

concrete solutions if we only need to

know how many solutions exist . Besides

for large numbers, for example 100 instead

of 20, {it}^{'} s not easy to list all the possible

solutions .

We need to know the number of solutions for

n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20

under the condition

1 ⩽ n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

if we replace

n_{1} = a_{1} + 1

n_{2} = a_{2} + 2

n_{3} = a_{3} + 3

n_{4} = a_{4} + 4

n_{5} = a_{5} + 5

then

the equation n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20

will change to a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 5

and

the condiction 1 ⩽ n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

will change to 0 ⩽ a_{1} ⩽ a_{2} ⩽ a_{3} ⩽ a_{4} ⩽ a_{5}

but

the number of solutions of

a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 5

under the condition

0 ⩽ a_{1} ⩽ a_{2} ⩽ a_{3} ⩽ a_{4} ⩽ a_{5}

means the number of partitions of

the integer 5 which is denoted by

P (5) .

P (5) = 7 \Rightarrow number of solutions is 7 .

the 7 partitions of the integer 5 are :

a_{1} + a_{2} + a_{3} + a_{4} + a_{5} \Rightarrow n_{1} n_{2} n_{3} n_{4} n_{5}

0 0 0 0 5 1 2 3 4 10

0 0 0 1 4 1 2 3 5 9

0 0 0 2 3 1 2 3 6 8

0 0 1 1 3 1 2 4 5 8

0 0 1 2 2 1 2 4 6 7

0 1 1 1 2 1 3 4 5 7

1 1 1 1 1 2 3 4 5 6

Commented by Tinkutara last updated on 08/Oct/17

Is there any formula for partitions ?

How to list P (85) ?

Commented by mrW1 last updated on 09/Oct/17

there is no formula to calculate the

integer partitions directly . usually we

use generating functions .

Commented by Tinkutara last updated on 09/Oct/17

Got it .

Commented by mrW1 last updated on 09/Oct/17

mr . ajfour had an example in Q 21802

for using generating function to find

number of integer solutions . it is not

really a function but a method .

Commented by mrW1 last updated on 09/Oct/17

see also Q 21800

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