Permutation-Combination
Aptitude

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Q.

Goldenrod and No Hope are in a horse race with 6 contestants.

How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

 A.

$700$

 B.

$360$

 C.

$120$

 D.

$24$

 E.

$21$

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Solution:
Option(B) is correct

Two horses $A$ and $B$, in a race of 6 horses... $A$ has to finish before $B$

If $A$ finishes $1^{st}$... $B$ could be in any of other 5 positions in 5 ways and other horses finish in $4!$ Ways, so total ways, $= 5×4!$

If $A$ finishes $2^{nd}$... $B$ could be in any of the last 4 positions in 4 ways. But the other positions could be filled in $4$! ways, so the total ways, $= 4×4!$

If $A$ finishes $3^{rd}$... $B$ could be in any of last 3 positions in 3 ways, but the other positions could be filled in $4!$ ways, so total ways, $= 3×4!$

If $A$ finishes $4^{th}$... $B$ could be in any of last 2 positions in 2 ways, but the other positions could be filled in $4!$ ways, so total ways, $= 2 × 4!$

If $A$ finishes $5^{th}$... $B$ has to be 6th and the top 4 positions could be filled in $4!$ ways.

$A$ cannot finish $6^{th}$, since he has to be ahead of $B$.

Therefore total number of ways:

$=5×4! + 4×4! + 3×4! + 2×4! + 4!$

$= 120 + 96 + 72 + 48 + 24$

$=\textbf{360}$


(1) Comment(s)


Abcd
 ()

couldn't the explanation just be total combination 6! , but in half of them A will be before B and other half B will finish before A therefore answer is

6! / 2.