Question

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 numbers of such distinct arrangements $({n_1}.{n_2}.{n_3}.{n_4}.{n_5})$ is…………………………

Answer 1

Hint: In this question we have to find the all the distinct positive integers arrangements of the $({n_1}.{n_2}.{n_3}.{n_4}.{n_5})$ and we two condition given in the question which is ${n_1} < {n_2} < {n_3} < {n_4} < {n_5}$,${n_1} + {n_2} + {n_3} + {n_4} + {n_5} = 20$. we can solve this question by using the combination method.

Complete step-by-step answer:
For solving these questions we have to take the ${n_5}$ value from 10 to 6 and the carry forward moves from 0 to 4 because the difference between the 10 and 6 is 4 . first make the arrangement using the numbers
When we arranged the ${n_5}$ carry forward moves from 0 to 4
And the arrangement looks like
$^4{C_0} + \dfrac{{^4{C_1}}}{4} + \dfrac{{^4{C_2}}}{3} + \dfrac{{^4{C_3}}}{2} + \dfrac{{^4{C_4}}}{1}$
Here we have to use the combination method that is $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
When we solve the above functions, we get
$^4{C_0} = \dfrac{{4!}}{{0!(4 - 0)!}}$
$^4{C_0} = \dfrac{{4.3.2.1}}{{0!.4.3.2.1}}$$ = 1$
In the above equation 0! Is equal to 1 and the denominator is cancelled by nominator$^4{C_1} = \dfrac{{4!}}{{1!(4 - 1)!}}$
$\dfrac{{^4{C_1}}}{4} = \dfrac{{\dfrac{{4.3.2.1}}{{1.3.2.1}}}}{4}$
$ = \dfrac{4}{4} = 1$
Third
$^4{C_2} = \dfrac{{4!}}{{2!(4 - 2)!}}$
$^4{C_2} = \dfrac{{\dfrac{{4.3.2.1}}{{2.1.2.1}}}}{3}$
$ = \dfrac{6}{3} = 2$
Fourth
$^4{C_3} = \dfrac{{4!}}{{3!(4 - 3)!}}$
$^4{C_3} = \dfrac{{\dfrac{{4.3.2.1}}{{3.2.1.1}}}}{2}$
$ = \dfrac{4}{2} = 2$
Fifth
$^4{C_4} = \dfrac{{4!}}{{4!(4 - 4)!}}$
$^4{C_4} = \dfrac{{\dfrac{{4.3.2.1}}{{4.3.2.1.0!}}}}{1}$
Multiply the numbers in denominator and denominator.
$ = \dfrac{4}{4} = 1$
Dividing the nominator by denominator
After adding all we get
$^4{C_0} + \dfrac{{^4{C_1}}}{4} + \dfrac{{^4{C_2}}}{3} + \dfrac{{^4{C_3}}}{2} + \dfrac{{^4{C_4}}}{1}$
=1+1+2+2+1
Add the numbers
= 7

So, the number of distinct arrangements is 7.

Note: Use the combination method to solve this question that is $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$. And always remember that 0! Is 1.this question is related to agreement. Student may do the mistake while doing the calculations here you can also do this question by possible ways as the question says that the sum of all the positive integers are 20 and every first integer is less then second so, the possible ways are
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
1,3,4,5,7
2,3,4,5,6
Here you can check that the sum of all the arrangements is 20 according to question.