Number of $3$ letter that can be formed from the letters of $A I E E E$ is $x$ and maximum number of intersection part can be obtained from $5$ different parabola is $y$ then $x + y =$

33

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

47

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

53

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

37

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

Open in App

Solution

The correct option is A $47$
$N u m b e r o f t h r e e l e t t e r f o r m e d f r o m A I R E E i s w h e n A I E i s t a k e n w e g e t 3 c o m b i n a t i o n w h e n A E E i s t a k e n w e g e t \frac{3}{2} c o m b i n a t i o n w h e n I E E i s t a k e n w e g e t \frac{3}{2} c o m b i n a t i o n a n d o n e c o m b i n a t i o n o f E E E a n d 2^{5} c r o s s sec t i o n p o i n t f o r p a r a b o l a$

So, $15 + 32 = 47$

Hence, this is the answer.

Suggest Corrections

Similar questions

X

Y

Z

_{19}^{40} X

_{19}^{39} Y

_{20}^{40} Z

X

Y

Z

6

S U M A N

D I V Y A

3

N^{2}

N

Find the number of different 4-letter words, with or without meanings, that can be formed from the letters of the word 'NUMBER'.

Join BYJU'S Learning Program