it is 38

how...

maihuna wrote:24: if you have one white, two blue, and three red flags, find how many different signals can be made, each containing four colors arranged one above the other.

19
28
38
120
360

1w, 2B, 3R choose any 4.

Case1: 1w, 1B,2R. possible permutation is 4!/2!=12
case2: 1W,2B, 1R i.e 4!/2!=12
case3:1W, 3R= 4!/3!=4
Cae4: 2B, 2R= 4!/2!2!=6
case5:1B, 3R= 4!/3!=4

total ways : 12+12+4+6+4=38ways

I am getting an answer not in the options..can someone help me

4 colors out of 6= 6C4=15 ways
these 4 colors can be in 3 different ways:
1,1,2
2,2
1,3

1,1,2:
the 4 colors can be arranged in 15*4!/2! ways= 180

2,2:
the 4 colors can be arranged in 15*4!/(2!*2!) ways= 90 ways

1,3:
the 4 colors can be arranged in 15*4!/(3!) ways=60 ways
total=330 ways
where am I wrong??