I am unable to see video . i can just hear audio . is there any issue with this video ? or its frm my side ?
Its working properly.
Sir in palindroms problem Task 5,6,7 dependent on Task 1,2,3 respectively.then how can we apply here Product rule?
What ever the way T1 is done if T2 has equal no of ways then we say tasks T1,T2 are independent.
Here for every 26 ways of filling 1st position we have equal, that is one, no of ways of filling last position
can we say that T7 is dependent on T1?
sir for 5 digit integers problem we should use 6 positions as 1st position for sign(negative or positive) and 2nd-6th for 5 digit number.
in that way the ans will be 2x9x10^4.
this is related to 2nd lecture. by mistake added here.
In the question I was discussing positive integers
sir according to product rule at 8:24 possblie solution would be 5*9*8*7*6 ie 5 for raju snd remaing 4 for other persons but sir according to the combination 5* 9C4=(9*8*7*6*5)/4! answer are different
I understand your problem. Can you tell me for which problem you get these 2 solutions ?
@gaurav27pro Please elaborate your probem…
As it is 5 member committee . and by looking at your number of people are 10
If raju sits at a fixed seat or position leaving other 4 seats or position for the remaining members then sitting of raju can be done only in 1 way i.e.
answer should be (1*9*8*7*6) by product rule.
And according to definition of combination you to select people first((9C4) and then arrange them(4!) later , assuming raju sits on a fixed position. Then total number of ways of committee forming is 1*( 9C4 * 4! ) which is equal to (1*9*8*7*6)
Sir , I can’t solve this question and my ans showing wrong .
Please watch all the combinatorics lectures and then try to answer this question. After that if you can not answer this question then please post your question in the below forum. http://forum.thegatebook.in/
4*9^3(4 mean fill 5 on 4 places )
Sir, C1(1,2,0) and C2(0,2,0). Sir there can be Case (1,1) when both rani and vani will go and when only vani will go(0,1). Sir,these two can be considered or not?
yes, they are considered in the 3*3 possibilities.
link for practising combinatorics question??
At 24:40 why taking t5 as 4 and t6 as 3….why we cannot select as 4C2 ways for selecting 2 places for 5 and 6