How many terms are in the expansion of $(x_{1} + x_{2} + x_{3})^{4}$

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $15$
Number of terms $= N =^{n + r - 1} C_{r - 1}$
$N =^{4 + 3 - 1} C_{3 - 1} =^{6} C_{2}$
$N = \frac{6!}{(6 - 2)! 2!} N = \frac{6!}{4! 2!} = \frac{6 \times 5 \times 4!}{4! \times 2} = 15$

Suggest Corrections

Similar questions

{(x_{1} + x_{2} + \dots . + x_{p})}_{1}^{n}

- 15

- 12

- 9

\dots \dots

How many terms of the series 18 + 15 + 12 + ...... when added together will give 45 ?

18 + 15 + 12 + . . . . . . .

45

N = 12^{12} \times 14^{14} \times 15^{15}

K = 12^{10} \times 14^{10} \times 15^{10}

Join BYJU'S Learning Program