Permutation And Combination - Study Mode
[#136] There are 7 non-collinear points. How many triangles can be drawn by joining these points?
Correct Answer
(A) 35
Explanation
Solution: A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points.
The number of triangles formed, $$eqalign{
& { = ^7}{{ ext{C}}_3} cr
& = frac{{7 imes left( {7 - 1}
ight) imes left( {7 - 2}
ight)}}{{3!}} cr
& = frac{{7 imes 6 imes 5}}{{3 imes 2 imes 1}} cr
& = 7 imes 5 cr
& = 35 cr} $$
[#137] From 6 men and 4 ladies, a committee of 5 is to be formed. In how many ways can this be done, if the committee is to include at least one lady?
Correct Answer
(A) 246
Explanation
Solution: To committee can be formed in the following ways, (1 lady + 4 gents) or (2 ladies + 3 gents) or (3 ladies + 2 gents) or (4 ladies + 1 gents) or (5 ladies + 0 gents). Total number of possible arrangements, = ( 4 C 1 × 6 C 4 ) + ( 4 C 2 × 6 C 3 ) + ( 4 C 3 × 6 C 2 ) + ( 4 C 4 × 6 C 1 ) = 60 + 120 + 60 + 6 = 246
[#138] The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is:
Correct Answer
(D) 10
Explanation
Solution: If n is even, then the number of boys should be equal to number of girls, let each be a .
⇒ n = 2a Then the number of arrangements = 2 × a! × a! If one more students is added, then number of arrangements, = a! × (a + 1)! But this is 200% more than the earlier ⇒ 3 × (2 × a! × a!) = a! × (a + 1)! ⇒ a + 1 = 6 and a = 5 ⇒ n = 10 But if n is odd, then number of arrangements,
= a!(a + 1)! Where, n = 2a + 1 When one student is included, number of arrangements, = 2(a + 1)! (a + 1)!
By the given condition, 2(a + 1) = 3, which is not possible.
[#139] How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4 if repetition of digits is allowed?
Correct Answer
(D) 376
Explanation
Solution: The smallest number in the series is 1000, a 4-digit number. The largest number in the series is 4000, the only 4-digit number to start with 4. The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3. The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4. Hence, there are 3 × 5 × 5 × 5 or 375 numbers from 1000 to 3999 Including 4000, there will be 376 such numbers.
[#140] How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated.
Correct Answer
(C) 216
Explanation
Solution: Test of divisibility for 3: The sum of the digits of any number that is divisible by 3 is divisible by 3 For instance, take the number 54372 Sum of its digits is 5 + 4 + 3 + 7 + 2 = 21 As 21 is divisible by 3 , 54372 is also divisible by 3 There are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits. The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by 3 Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either 0 or 3 while forming the five digit numbers. Case 1 If we do not use '0', then the remaining 5 digits can be arranged in: 5! ways = 120 numbers. Case 2 If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'. The first digit from the left can be any of the 4 digits 1, 2, 4 or 5 Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways. So, there will be 4 × 4! numbers = 4 × 24 = 96 numbers. Combining Case 1 and Case 2, there are a total of 120 + 96 = 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.