1.The total no. of all proper factors of 75600 is (ans- 118)
2. If nC4,nC5,nC6 are in AP, then the value of n is (ans- 7 & 14)
3. Let the exponent of 7 in 100C50 is t. Then t will be
(a) divisible by 7
(b) div by 100
(c) div by 50
(d) none
1ans 1) 75600=24.33.52.71
so the total number of factors of the given no. will be (4+1)(3+1)(2+1)(1+1)=120
but this contains the numbers 1 and 75600 itself so this won't be there in the ans
ans=120-2=118
i want others to try 2 and 3 as they r very straightforward
106Q1. 75600 = 33.71.52.24
so total no of divisors = (3+1)(1+1)(2+1)(4+1) = 120
so no. of prper divisors = 120-2 = 118
Q2. 2nC5 = nC4 + nC6
=> 2/5!(n-5)! = 1/4!(n-4)! + 1/6!(n-6)!
=> 2(n-4).6 = 5*6 + (n-5)(n-4)
=> 12n - 48 = 30 + n2 - 9n + 20
=> n2 - 21n + 98 = 0
=> n= 7,14
Q3. exponent of 7 in 100C50 = exp of 7 in 100! - 2*(exponent of 7 in 50!) = (14+2) - 2(7+1) = 0
so ... abc ..
