This is what i did in the exam

Why so much work? The number of sequences of 8 nonnegative integers whose sum is 4 is equal to $\binom{8 + 4 - 1}{4}$ (this is the usual stars-and-bars technique for counting compositions of an integer). These represent integers unless they begin with 0; the number that begin with 0 is the number of sequences of 7 nonnegative integers whose sum is 4, and this is equal to $\binom{7 + 4 - 1}{4}$. So subtract.

hmmm.... why subtraction. $ \binom{10}{3} $ is enough.

ayush_2008 wrote: hmmm.... why subtraction. $ \binom{10}{3} $ is enough. Because an answer is simply not enough. A solution, on the other hand, explains how the problem is tackled and JBL has shown a good way of doing it. Besides, $\binom{11}{4}-\binom{10}{4}=\binom{10}{3}$ to demystify the answer.

10000th User wrote: ayush_2008 wrote: hmmm.... why subtraction. $ \binom{10}{3} $ is enough. Because an answer is simply not enough. A solution, on the other hand, explains how the problem is tackled and JBL has shown a good way of doing it. Besides, $\binom{11}{4}-\binom{10}{4}=\binom{10}{3}$ to demystify the answer. if logic is there then whats the need of calculation i just give one star to first place then distribute three stars and 7 bars in seven places which are left.

Do you understand that writing down the number $\binom{10}3$ is not the same as writing down the combinatorial argument? The latter (which you did in your second post) is valuable (and, indeed, this is a quite direct way to solve the problem); the former (which is what you did in your first post) is totally useless. Or, to put it another way: a bit of logic does simplify the calculation, as you say. But your first post (to which 10000 was responding) doesn't contain any logic.

To clarify things up a bit:-- It can also be solved using multinomial(that's the name used in FIITJEE so don't know if it's correct) The question says to find the number of non $negative$ $integral$ solutions of the equation $x_1 + x_2 + x_3 + x_4 + ..... + x_8 = 4$ It's well known that the number of non negative integral solutions of $x_1 + x_2 + .... + x_n = r$ is $\binom{n+r-1}{n-1}$ so the number of solutions is $\binom{8+4-1}{7}$ However some numbers start with 0 hence we've to subtract such numbers .. The number of such numbers is $\binom{7+4-1}{7-1}$.. Hence Answer$=$ $\binom{8+4-1}{7}$ $-$ $\binom{7+4-1}{6}$. $=$ $\binom{10}{3}$ QED

Hi, this can also be solved with a few cases.

ayush_2008 wrote: 10000th User wrote: ayush_2008 wrote: hmmm.... why subtraction. $ \binom{10}{3} $ is enough. Because an answer is simply not enough. A solution, on the other hand, explains how the problem is tackled and JBL has shown a good way of doing it. Besides, $\binom{11}{4}-\binom{10}{4}=\binom{10}{3}$ to demystify the answer. if logic is there then whats the need of calculation i just give one star to first place then distribute three stars and 7 bars in seven places which are left. No, this is wrong .. If you assign the first place's value to be $1$ (which is perfectly fine, and needed) and distribute the three remaining $1's$ in the remaining $7$ places, you will miss all those cases where the first place is $2$ or $3$ or $4$.. I will provide a correct solution below..

While $\binom{10}{3}$ is the correct answer, I don't know why there is so much confusion to think simple.. Here's how you do it - We have got $8$ empty boxes to be filled up with four $1's$.. Now we know, the first box must have atleast one $1$, or else it will be $0$ and the whole number will not have $8$ digits.. So, why not just give a $1$ to the first place and satisfy it?.. Now that the first box is already fed with a $1$, we are sure that the number has exactly $8$ digits.. Therefore, the remaining three $1's$ will be distributed in these $8$ boxes, such that boxes may remain empty.. That's a very well known stars and bars method, and therefore the answer must be nothing but - $$\binom{8+3-1}{8-1}=\binom{10}{7}$$$$=\binom{10}{3}$$$$=\boxed{\boxed{120}}$$

$\blacksquare$