Jack

10-10-2013, 11:33 AM

In how many different ways can 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes be placed in a row?

(A) 48

(B) 576

(C) 1200

(D) 69300

Thanks

(A) 48

(B) 576

(C) 1200

(D) 69300

Thanks

View Full Version : sat math probability permutations

Jack

10-10-2013, 11:33 AM

In how many different ways can 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes be placed in a row?

(A) 48

(B) 576

(C) 1200

(D) 69300

Thanks

(A) 48

(B) 576

(C) 1200

(D) 69300

Thanks

miranda

10-10-2013, 11:34 AM

In this case we have permutations since the order matters. Additionally we must divide the number of permutations by 2! , 4!, 3! and 2! Since there are 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes.

Therefore the answer is \frac{^11P_{11}}{(2!) \cdot(4!) \cdot(3!) \cdot(2!)}= \frac{11!}{576}=69300 (D).

Therefore the answer is \frac{^11P_{11}}{(2!) \cdot(4!) \cdot(3!) \cdot(2!)}= \frac{11!}{576}=69300 (D).

Jack

10-10-2013, 11:39 AM

Thank you!!

Powered by vBulletin® Version 4.2.2 Copyright © 2017 vBulletin Solutions, Inc. All rights reserved.