# pr.probability – Doubt in Question

Question: How many different letter permutations, of any length, can be made using the letters M O T T O. (For instance, there are 3 possible permutations of length 1.)

Hey guys, I was busy attempting this problem, but I am confused on in 3 where XXYZ in 2C1 × 4! ÷ 2! why do we have to multiply by 2C1. I understand the reasoning behind everything else except for this. Thanks.

1
Three ways i.e. {M,T,O}

2
XX or XY
XX in 2C1 = two ways i.e. {OO or TT}
XY in 3C2 × 2! = six ways

3
XXY or XYZ
XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve ways
XYZ in 3C3 × 3! = six ways

4
XXYY or XXYZ
XXYY = 4! ÷ (2! × 2!) = six ways
XXYZ in 2C1 × 4! ÷ 2! = twenty four ways

5
= 5! ÷ (2! ×2!)
= 120 ÷ 4
= 30
Therefore, the total is
= 3 + (2 +6)+ (12 +6) + (6 +24) + 30
= 89