@@brucebaelon Hindi po yan repeated. Ang first option natin is 4 numners lang. Ang 2nd ay 5 numbers because sa anim nabawasan na ng isa sa unahan, so is the 3rd. Kaya 4 numbers nalang kasi 6-2=4. Yung 6 ay every number while ang 2 ay ang nakuha na na number of digits.
Dr Anil Kuma, For question (c), there is another alternative as following: For question (b) above Odd numbers + Even numbers = 240 ways ==> Even numbers = 240/2 = 120 ways
An easier version of letter C "Even numbers without repitition". 4 digits greater than 3000 from 1,2,3,4,5,6 1. Count the number of even numbers among 1,2,3,4,5,6 that is greater than 3000. That will be {4,6} or 2 ways so our first number will be 2 ,_ ,_ ,_ 2. Since in the first number we could choose either 4 or 6, that means we still have 5 more numbers to choose for the 2nd as repition is not allowed. if we choose 2, then we still have 6,5,4,3,1 to choose from or 5 ways if we choose 6, then we still have 6,5,3,2,1 to choose from or 5 ways that means our next number is 5. 2,5,_ ,_ (Just continure to use step 2 for the remaining numbers) Then 4 options to choose for the third, and 3 options to choose for the last we now have 2,5,4,3 2,5,4,3 = 120 ways I'm not sure if this is correct but this is how i've done it before.
I have resit exams honestly with covid 19 occupy my mind I haven’t understood nothing so I cross my fingers to overcome this and thanks for little explanation
Thanks a lot for that! Highly appreciate. Earlier generation was communicating with letters. I remember so may paersons whose handwritten notes could match any printed book.
Concepts are very helpful sir but i face difficulty when the same sum comes like what will be the summation of all nos greater than n formed by dihits like a b c d. Plz make a video on this problem too.
Sir, what if they didn't mention anything about repetition in the questions?? Like: how many 4digit no. greater than 7000 can be formed out of the digits 3,5,7,8,9
Thank you for the response, Sir. But which method(Repetition or without repetition) should I follow if they didn't give any specifications in the question?
Sir, Can you plz answer thiz.. Consider all permutations of the 16 numbers from 1 to 16 which satisfy the property that every number is placed such that it is either bigger than ALL numbers preceding it or it is smaller than ALL numbers preceding it. The number of such permutations is
I don't think your c is right.....please recheck because in the 2nd case you're not supposed to multiply 4 at the end rather 1 since 4 is already a fixed number which can only be chosen in 1 way......sorry yours is right 😂...I paused the video to right this comment right before you spotted your mistake
suk bahadur 3,4,5,6 can placed in thousands Once one of this is taken you are left with 5 numbers Then 4 That is how we get our solution Hope that helps
I don't understand why 6is fixed in the last one, units place; If there needs to be an even number, shouldn't all three of them from the given numbers be considered?
Munda Punjabi , it’s quite easy to get confused with permutation and combination specially when we allow the repetition in permutation. Consider this example, I have to make a fruit salad and I have 3 items; banana,apple and mango. would the order really matter in the condition? Eventually, it will be a fruit salad, it won’t matter which fruit I choose first. Now consider this one let’s say I have to set a password for my phone which is 6789, if i change the order, assume 7689, would my phone get unlocked?. In this example order really matters. Just practice things out, you will get a sense of it
What happens when the number begins with 3 then followed by three repeated 0's? That becomes 3000 of which it's not supposed to be included in (a). Oops.......... Asking for a roomate
Please rewrite the question with all the options. 3 digit number without repetition (number should not start with zero): 9 x 9 x 8 should be the answer. Hope that works. Thanks
