Let A=(a,b,c):a,b,c are non-negative integers and a+b+2c=22.

Question

Let

$A={(a,b,c):a,b,c$ are non-negative integers and $a+b+2c=22}$.

Then $n(A)$ is equal to :

Accepted Answer

We need to count the number of ordered triples $(a,b,c)$ of non-negative integers satisfying

$a+b+2c=22$

Since $a,b,c\ge 0$, let us fix $c$ first.

For a fixed value of $c$, the equation becomes

$a+b=22-2c$

Now, the number of non-negative integer solutions of

$a+b=n$

is

$n+1$

So, for fixed $c$, number of solutions is

$(22-2c)+1=23-2c$

Now find all possible values of $c$.

Since $2c\le 22$,

$c=0,1,2,\ldots ,11$

Hence total number of solutions is

$\sum _{11}^{c=0}(23-2c)$

This is an arithmetic series:

first term $=23$

last term $=23-2(11)=1$

number of terms $=12$

So,

$n(A)=\frac{12}{2}(23+1)$

$n(A)=6\cdot 24=144$

Therefore,

$n(A)=144$

So the correct option is:

$144$

Hence, .

Solutions · 1