0

Today I had my exam of discrete maths and was asked to find the:

no. of $10$-combinations of multiset $\{\infty a,3b,6c\}$.

What I did was that:

consider set $A_1=$ no. of ways such that no element of $10$ go in $a$ ,
similarly consider $A_2=$ no. of ways such that atleast $4$ elements go in $b$ and
$A_3=$ no. of ways such that atleast $7$ of $10$ elements go in $c$ .

then find required ways as=$|S|-(|A_1|+|A_2|+|A_3|)+(|A_1\cap A_2|+|A_2\cap A_3|+|A_3\cap A_1|)-(|A_1\cap A_2\cap A_3|)$

using inclusion-exclusion principle....

Am I correct doing this?

JacKeown
  • 260
patang
  • 797

1 Answers1

1

The infinity a symbol represents that a can occur in the combination of 10 as many times as you like. So, there would be no A1 as there is no restriction on the number of a that can occur in the combination.

Otherwise, your use of the PIE is correct, except that intersection of A2 and A3 is empty because you cannot have 4b and 7c together. So the total number of combinations is simply |S| - |A2| - |A3|.

Sean Lo
  • 437