Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The number of distinct terms in the expansion of  x1+x2++xn3 is

 

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

a

 n+1C3

b

 n+2C3

c

 n+3C3

d

 nC3

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

We know that

x1+x2++xn3=3!λ1!λ2!λn!x1λ1x2λ2xnλn

where λ1+λ2++λn=3 and λi0

Thus, number of distinct terms in the expansion of

x1+x2++xn3

is equal to the number of non-nagative integral solution of 

λ1+λ2++λn=3

= number of ways of distributing 3 identical objects among n persons

=3+n1C3=n+2C3 

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring