let n be the number of 8 digit numbers the sum of whose digit is 4 . find n/12
Answers
Step-by-step explanation:
Correct option is
A
120
We need to find the number of 8-tuples ( a
1
,a
2
,...,a
8
) of non-negative integers such that a
1
≥1 and a
1
+a
2
+...,+a
8
= 4. If a
1
=1.
There are three possibilities :
Either exactly three among a
2
,a
3
,...,a
7
equal 1 and the rest equal zero,
or
Five of them are zero and the other two equal 1 and 2,
or
Six of them are zero and the other equals 3.
In the rest case, there are (
7
3
) = 35 such 8-tuples,
In the second case there are (
7
2
) = 42 such 8-tuples
and in the third case there are 7 such 8-tuples.
If a
1
=2 then either six of a
2
,a
3
,..,a
7
are zero and the other equals two, or five of them are zero and the remaining two both equal 1.
In the former case, there are 7 such 8-tuples and in the latter case there are (
7
2
) = 21 such 8-tuples.
If a
1
=3 then exactly six of a
2
,a3,...,a
7
are zero and the other equals one.
∴ 7 such 8-tuples.
Finally, there is one 8-tuple in which a
1
=4 .
∴ there are 120 such 8-tuples
so n/12 is equal to 120/12 = 10