@Shray14 said:There are 6 person with 6 tasks. Task 1 cannot be assigned to either 1 or 2 . Task 2 must be assigned to either 3 or 4..every person must assigned one task. In how many ways does assignement be done??
144 ways?
@Shray14 said:There are 6 person with 6 tasks. Task 1 cannot be assigned to either 1 or 2 . Task 2 must be assigned to either 3 or 4..every person must assigned one task. In how many ways does assignement be done??
144????
@Shray14 said:Correct..Approach??
first assign task 2 in 2 ways(3 or 4) then task 1 can be assigned in 3 ways..
Remaining tasks in 4! ways..
Total 2*3*24 = 144
@SarayuSheshadri said:first assign task 2 in 2 ways(3 or 4) then task 1 can be assigned in 3 ways..Remaining tasks in 4! ways..Total 2*3*24 = 144
How task 1 assign in 3 ways??
@Shray14 said:How task 1 assign in 3 ways??
Consider I assign task 2 to 3.. Now task 1 can be assigned to either 4,5 or 6..
Similarly if I assign task 2 to 4 task 1 can be assigned to 3,5 or 6..
@SarayuSheshadri said:Consider I assign task 2 to 3.. Now task 1 can be assigned to either 4,5 or 6..Similarly if I assign task 2 to 4 task 1 can be assigned to 3,5 or 6..
Ohkey...i was not including 4 after assigning task 2 to3.
similarly to 3 after task 2 to 4..
silly mistake..
Thanks bro.
@gnehagarg said:dearrangement of 5!=44
@LeoN88 said:@rubikmathAns. is 14 ?Approach is lengthy one..will post if the ans. is correct !
oa is 14
it doesnt matter if after rearrangement they are in original position . it is allowed .
@rubikmath said:@SarayuSheshadrioa is 14 it doesnt matter if after rearrangement they are in original position . it is allowed .
I know its allowed.. I was telling @gnehagarg that derangement is not the right approach in this case..
And guess I missed two arrangements somewhere :)
@SarayuSheshadri
the method i found in the source was
for N pots
(N-1)! - summation (k!)
where k = 0 to N-2
but duno how they got this ..
the method i found in the source was
for N pots
(N-1)! - summation (k!)
where k = 0 to N-2
but duno how they got this ..
5 people are travelling in a car . 1 father , 1 mother , 2 daughters , 1 son . There are 2 front seats and 3 back seats in the car . Either the father or the mother has to drive the car, and the daughers wont sit next to each other . possible arrangements ?
a. 8
b. 32
c. 40
d. 24
e. none
@rubikmath said:5 people are travelling in a car . 1 father , 1 mother , 2 daughters , 1 son . There are 2 front seats and 3 back seats in the car . Either the father or the mother has to drive the car, and the daughers wont sit next to each other . possible arrangements ?a. 8b. 32c. 40d. 24e. none
B. 32?
Consider father driving.
4 seats remaining. Total possible combinations 4!=24.
Cases where daughters sit together=(2!*2!) + (2!*2!) = 8
Remaining cases= 24-8= 16
Similar when mother driving.
Possible cases= 16*2 =32
Consider father driving.
4 seats remaining. Total possible combinations 4!=24.
Cases where daughters sit together=(2!*2!) + (2!*2!) = 8
Remaining cases= 24-8= 16
Similar when mother driving.
Possible cases= 16*2 =32
@rubikmath said:@SarayuSheshadrithe method i found in the source was for N pots(N-1)! - summation (k!)where k = 0 to N-2but duno how they got this
hmmm.. lets see.. If I figure out I'll post here..
Meanwhile the gods of the quant thread will help us out :)
@rubikmath said:5 people are travelling in a car . 1 father , 1 mother , 2 daughters , 1 son . There are 2 front seats and 3 back seats in the car . Either the father or the mother has to drive the car, and the daughers wont sit next to each other . possible arrangements ?a. 8b. 32c. 40d. 24e. none
32?
@rubikmath said:5 people are travelling in a car . 1 father , 1 mother , 2 daughters , 1 son . There are 2 front seats and 3 back seats in the car . Either the father or the mother has to drive the car, and the daughers wont sit next to each other . possible arrangements ?a. 8b. 32c. 40d. 24e. none
if both daughters are sitting at the back : 2*2*2 = 8 ways
if 1 daughter is in the front: 2*2*6 = 24 ways
total 32 cases..
If two circles of radius 3 cm and 5 cm intersect each other at two points and the distance between their centres is 4 cm. Then find the length of their common chord.
@rubikmath said:5 people are travelling in a car . 1 father , 1 mother , 2 daughters , 1 son . There are 2 front seats and 3 back seats in the car . Either the father or the mother has to drive the car, and the daughers wont sit next to each other . possible arrangements ?a. 8b. 32c. 40d. 24e. none
My take 32
Just fix the father in driver's position,then possible arrangements may be 3! + 3! + 2!*2!
And now if mother drives,then we have another 16 arrangements possible...
So,2*(3! + 3! + 2!*2!) = 32
@astute99 said:If two circles of radius 3 cm and 5 cm intersect each other at two points and the distance between their centres is 4 cm. Then find the length of their common chord.
6cm?