Question 21
Question: A letter lock consists of three rings, each marked with fifteen different letters. Find in how many ways it is possible to make an unsuccessful attempt to open the lock.
Answer: 1st ring can be attempted in 15 ways.
1st and 2nd ring can be attempted in 15 x 15 = 152 ways
1st and 2nd and 3d ring can be attempted in 15 x 15 x 15 = 153 ways
Among these 153 attempts, one attempt will be a successful attempt.
Hence the number of unsuccessful attempts is 153-1= 3375 -1 = 3374.
Question 22
Question: Find the sum of all four digit numbers formed by using 2, 3, 6, 9 in which no digit is repeated.
Answer: If 2 occupies unit's place, the remaining 3 digits can be arranged in 3!= 6ways. Similarly, if 2 occupies ten's place, hundred's place, thousand's place, in each of these cases we get 3! numbers. Thus, the positional value contributed by 2 to the sum when it occupies different
values is (3!)(2) + (3!)(20) + (3!)(200) + (3!)(2000) = 3!(2)(1111)
Similarly, the values contributed by 3, 6, 9 to the sum are
3! (3)(1111), 3! (6) (1111), 3! (9)(1111) respectively.
The required sum is 3!(1111)(2+3+6+9) = 1,33,320
Question 23
Question: Find n, r if
Answer: i. C(n,4) = 210
(
210 = 10 x 7 x 3)
ii. C(n,18) = C(n,12)
Using C(n,r) = C(n,n-r)
The value of C(32,n) = C(32,30) = C(32,2)
Question 24
Question: If the ratio C(2n,3):C(n,3) = 11:1, find n.
Answer: 
Question 25
Question: 
Answer: 
