Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 440 off
USE CODE: MOB20

```				   Let X={a1,a2......a6} and Y={b1,b2,b3}.Find number of functions f from x to y such that it is onto and there are exactly three elements x in X such that  f(x)=b1.
```

7 years ago

Share

### Answers : (1)

```										In mathematics, a function f is said to be surjective or onto, if its values span its whole codomain; that is, for every y in the codomain, there is at least one x in the domain such that f(x) = y .
We can choose any 3 of a1,a2....,a6, say c1,c2,c3. f(c1)=f(c2)=f(c3)=b1.
[For example, lets say f(a2),f(a5),f(a6) are equal to b1. Then, c1,c2,c3 are a2,a5 and a6]
c1,c2 and c3 can be chosen in 6C3 = 20 ways.
After removing c1,c2 and c3 from a1,a2,....a6, we will have three more a's left. Let these be d1,d2 and d3.
[In the above example, d1,d2 and d3 will be a1,a3, and a4]
We need f(d1),f(d2) and f(d3) to be either b2 or b3, but at least one of them should be b2 and at least one of them should be b3 because the function f is onto.
So the possible ways we can do this = 2*2*2 - 2 = 6
[Because we can assign f(d1) as b2 or b3. There are 2 ways of doing this. Similarly, we can assign f(d2)=b2 or b3 and same for f(d3). There are 2*2*2=8 ways of doing this. Then we subtract the two cases where all three are b1 or b2]
So total number of such functions = 20 * 6 = 120.
Let me know if you face any problem in understanding this explanation, especially if you have not yet done Permutations and combinations.

```
7 years ago

# Other Related Questions on Algebra

If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...

 Ajay 6 months ago

Small Mistake in last para posting again..............................................................................................................

 Ajay 6 months ago

We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.

 mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??

The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below

 Ajay 5 months ago

Thankyou so much............................. …......................................................................!

 Anshuman Mohanty 5 months ago

Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7

 Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above

If |z-i| = ?? PLs complete the question

 Nishant Vora one month ago

Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...

 Divya one month ago

I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above

 Divya one month ago
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent

Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...

 Kumar 3 months ago

Not any people get my answer why. You can no give answer my question I am join this site

 Vivek kumar 5 months ago

Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?

 Ajay 5 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2

What needs to be solved here ? The question is incomplete....................................................................

 Ajay 6 months ago

i don’t know how to do this...............................................................................................

 Saravanan 2 months ago

this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)

 Naveen Shankar 6 months ago
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee

Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H. The following can be proved easily: ​1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends...

 mycroft holmes one month ago

Draw which is Isoceles as OB = OC. Now which means . Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required...

 mycroft holmes one month ago

a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)

 mycroft holmes one month ago
View all Questions »

• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: R 15,000
• View Details
Get extra R 3,000 off
USE CODE: MOB20

Get extra R 440 off
USE CODE: MOB20

More Questions On Algebra

## Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details