Question

The number of functions ${\text{f}}$ from the set $A = \{ 0,1,2\} $ to the set $B = \{ 0,1,2,3,4,5,6,7\} $ such that $f(i) \leqslant f(j)$ for $i < j$ and $i,j \in A$ is
A. ${}^8{C_3}$
B. ${}^8{C_3} + 2({}^8{C_2})$
C. ${}^{10}{C_3}$
D. ${}^9{C_3}$

Answer 1

Hint: We can solve the above problem by knowing that if we need to select $n$ number of things from the total of $m$ number we can do it in ${}^m{C_n}$ ways. Here we are given the two sets, set $A = \{ 0,1,2\} $ to the set $B = \{ 0,1,2,3,4,5,6,7\} $
So here we need to make the cases when the function will satisfy the conditions$f(i) \leqslant f(j)$ for $i < j$ and$ i,j \in A$ . That number will give us our total count.

Complete step-by-step answer:
Here we need to find the number of functions ${\text{f}}$ from the set $A = \{ 0,1,2\} $ to the set $B = \{ 0,1,2,3,4,5,6,7\} $ such that $f(i) \leqslant f(j)$ for $i < j$ and $i,j \in A$
As we know that $A = \{ 0,1,2\} $ so we come to know that $i,j \in \{ 1,2\} $
As we are given $f(i) \leqslant f(j)$ so we need to arrange codomain in the ascending order.
For example: If we have the function from set A to set B as $\{ (1,2),(2,3),(3,4)\} $ so here we see that the in the first bracket $2 > 1$ and similarly in the second and third we get that $3 > 2,4 > 3$
Hence similarly here also the codomain must be greater than the element itself.
Now we can make the cases of selecting the value from the codomain.
Case (1)
If we select one element from the set B for the three elements of the set A we get the number of ways as:
Here we need to select one form those $8$ elements so we can do it in ${}^8{C_1}$ ways.
Case (2)
If we take three numbers from the set B for the three elements for the set A and arrange the elements of the set B in the ascending order we will get the number of ways:
Here we need to select $3$ out of $8$ and hence we get ${}^8{C_3}$ ways.
Case (3)
Here if we take the two same elements from the set B then we can get the number of ways:
We can arrange it in two ways which are like:
For the elements like we have for $0,1$ we can have the same value and in the second case we can have that the value are same for $1,2$ and differ for zero.
Hence selecting two elements out of $8$ can be done in ${}^8{C_2}$ ways but here there are two cases so we can do it in $2({}^8{C_2})$ ways.
Hence now we can do the sum which gives us
${}^8{C_1} + {}^8{C_3} + 2.{}^8{C_2}$
$8 + \dfrac{{8!}}{{3!(8 - 3)!}} + 2.\dfrac{{8!}}{{2!(8 - 2)!}}$
Here we must know the formula that
${}^m{C_n} = \dfrac{{m!}}{{n!(m - n)!}}$
Now we get that
$
  8 + \dfrac{{8!}}{{3!(8 - 3)!}} + 2.\dfrac{{8!}}{{2!(8 - 2)!}} \\
   = 8 + \dfrac{{8.7.6}}{6} + \dfrac{{2.8.7}}{2} = 120 \\
 $
Hence we get that there will be $120$ ways.
Now we need to check the options numerical value.
Value of
$
  {}^8{C_3} = \dfrac{{8.7.6}}{6} = 56 \ne 120 \\
  {}^{10}{C_3} = \dfrac{{10.9.8}}{6} = 120 \\
 $
Hence we get option C as correct.

Note: In solving these kind of question we must avoid the calculation errors and we need to solve it by knowing the proper formula which are like${}^m{C_n} = \dfrac{{m!}}{{n!(m - n)!}}$. We can solve the above problem by knowing that if we need to select $n$ number of things from the total of $m$ number we can do it in ${}^m{C_n}$ ways.