Number of ways in which 4 letter words can formed by using the letters from the set-a,a,a,a,b,b,b,c,c,d

Question

I thought of solving this question by making cases:

1) All four are same

2) 2 are same,2 are distinct

3) 3 are same ,1 is distinct

4) All are distinct

My difficulty is about how to select identical numbers

Like for case 2.How will I choose between-a,a,a, b,b, and c,c for 2 same numbers

Can somebody tell me about this

Answer 1

The five partitions of $4$ are

• 1+1+1+1,
• 2+1+1,
• 2+2,
• 3+1, and
• 4.

So, if you want to do it this way, you're still missing a case. It can be done this way though, with careful bookkeeping.

Partition 1+1+1+1: We must have a, b, c, and d in some order: $4!=24$ ways.

Partition 2+1+1: We have two copies of one letter from a, b, or c, and choose two other letters to write uniquely, and once these letters are chosen we can write it in $4!/2$ ways (we divide by $2$ since if we count all $4!$ permutations, we count each twice when we swap the two identical letters): $3 \times \binom{3}{2} \times 4!/2=108$ ways.

Partition 2+2: We choose two letters from a, b, or c, and we can write them in $\binom{4}{2}$ ways: total $\binom{3}{2} \times \binom{4}{2}=18$ ways.

Partition 3+1: We have three copies of one letter from a or b, and choose one other letter to write uniquely, and once the letters are chosen, we can write it in $4$ ways: $2 \times \binom{3}{1} \times 4=24$ ways.

Partition 4: We write aaaa: $1$ way.

Totaling: $$ 24+108+18+24+1=175 $$ ways.

We can check this by generating them with the GAP code:

Q:=Filtered(Tuples([1..4],4),T->ForAll([1..4],i->Number(T,t->t=i)<=i));

And Size(Q); returns 175.