For a fixed positive integer n, if D=n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!then Dn!(n+1)!(n+2)! is equal to

We can write D asD=n!(n+1)!(n+2)!1 n+1 (n+1)(n+2)1 n+2 (n+3)(n+2)1 n+3 (n+4)(n+3)Applying R3→R3−R3,R2→R2−R1, we getD=n!(n+1)!(n+2)!1n+1(n+1)(n+2)012(n+2)012(n+3)⇒ Dn!(n+1)!(n+2)!=2

For a fixed positive integer n, if D=n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!then Dn!(n+1)!(n+2)! is equal to

We can write D asD=n!(n+1)!(n+2)!1 n+1 (n+1)(n+2)1 n+2 (n+3)(n+2)1 n+3 (n+4)(n+3)Applying R3→R3−R3,R2→R2−R1, we getD=n!(n+1)!(n+2)!1n+1(n+1)(n+2)012(n+2)012(n+3)⇒ Dn!(n+1)!(n+2)!=2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

For a fixed positive integer $n$ , if
$D = |\begin{array}{ccc} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{array}|$
then $\frac{D}{n! (n + 1)! (n + 2)!}$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We can write $D$ as

$D = n! (n + 1)! (n + 2)! |\begin{array}{lll} 1 n + 1 (n + 1) (n + 2) \\ 1 n + 2 (n + 3) (n + 2) \\ 1 n + 3 (n + 4) (n + 3) \end{array}|$

Applying $R_{3} \to R_{3} - R_{3}, R_{2} \to R_{2} - R_{1}$ , we get

$\begin{array}{l} D = n! (n + 1)! (n + 2)! |\begin{array}{ccc} 1 & n + 1 & (n + 1) (n + 2) \\ 0 & 1 & 2 (n + 2) \\ 0 & 1 & 2 (n + 3) \end{array}| \\ \Rightarrow \frac{D}{n! (n + 1)! (n + 2)!} = 2 \end{array}$