There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125
Official Answer is A.
Thanks Take Gmat team.
5!/3!=(5*4*3!)/3!=20
20 ways!…
A
good review question
Keep it simple!! 3 red cars can be arranged in 5C3 ways i.e 10. The balance 2 cars can be arranged in 2 ways( since there are now only 2 places vacant to display the blue and yellow car). Therefore total no. of ways=10*2=20
Permutation=5x4x3x2x1
Combination- Permutation/3! three are alike=20
got it:
the formula was:
Combination! : n!
so
5! : 3!
5x4x3x2x1 : 3x2x1 = 5×4 = 20
A – 20
dummy style:
11123
11231
12311
23111
11132
11321
13211
32111
21311
12131
11213
31211
13121
11312
21131
12113
31121
13112
21113
31112
formula :
still cracking
Answer is ‘A’.
‘A’ – 20 must be the answer.
We can use the formula for this type of PS = n!
—————–
(a!b!…z!)
where n – total number
a, b, …z are those entities which posses repetitions