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If nC4, nC5 and nC6 are in AP, then the value of n can be
The number of subsets of the set A = {a1, a2, …, an} which contain even number of elements is
Let Tn denote the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If Tn +1 – Tn = 21, then n equals
If no husband and wife play in the same game, then the number of ways in which a mixed double game can be arranged from among nine married couples is
Out of 9 men, 2 men can be chosen in 9C2 ways.
Since no husband and wife are to play in the same game, we have to select 2 women from the remaining 7 women.
This can be done in 7C2 ways.
If M1, M2 and W1, W2 are chosen, then a team can be constituted in 2 ways, viz. M1 W1 or M1 W2.
Thus, number of ways of arranging the game = 9C2 x 7C2 x 2 = 1512
The number of times the digit 3 will be written when listing the integers from 1 to 1000 is
In how many ways can 3 girls and 9 boys be seated in two vans, each having numbered seats with 3 in the front and 4 in the back, if 3 girls sit together in the back row on adjacent seats?
Total number of persons = 3 girls + 9 boys = 12
Total number of seats = 2 × 3 + 4 × 2 = 14 So, total number of ways in which 12 persons can be seated on 14 seats = Number of arrangements of 14 seats by taking 12 at a time = 14P12 Three girls can be seated together in the back row on adjacent seats in the following ways: 1, 2, 3 or 2, 3, 4 of first van And 1, 2, 3 or 2, 3, 4 of second van In each way, the three girls can interchange seats among themselves in 3! ways. So, total number of ways in which three girls can be seated together in the back row on adjacent seats = 4 × 3! Now, 9 boys are to be seated on remaining 11 seats, which can be done in 11P9 ways.
Hence, by the fundamental principle of counting, total number of seating arrangements = 11P9 × 4 × 3!
If n = 2015, then the least positive integer k for which k(n2) (n2 – 12) (n2 – 22) (n3 – 32) … (n2 – (n – 1)2) = r!, for some positive integer r, is
The number of ways in which we can arrange the digits 1, 2, 3, …, 9, such that the product of five digits at any of the five consecutive positions is divisible by 7, is
Let an arrangement of digits of a number be x1 x2 x3 x4 x5 x6 x7 x8 x9. Note that we require product of each of (x1, x2, x3, x4, x5); (x2, x3, x4, x5, x6); …; (x5, x6, x7, x8, x9) to be divisible by 7.
This is possible if the 5th digit is 7.
Therefore, we can arrange the 9 digits in desired number of ways,
Since position of 7 is fixed, the number of arrangements of remaining 8 digits is 8!.
An eight-digit number divisible by 9 is to be formed by using 8 digits out of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 without replacement. The number of ways in which this can be done is
We have 0 + 1 + 2 + 3 … + 8 + 9 = 45 To obtain an eight-digit number exactly divisible by 9, we must not use (0, 9) or (1, 8) or (2, 7) or (3, 6) or (4, 5). [Sum of the remaining eight digits is 36, which is exactly divisible by 9] When we do not use (0, 9), then the number of required 8 digit numbers is 8!. When one of (1, 8) or (2, 7) or (3, 6) or (4, 5) is not used, the remaining digits can be arranged in 8! - 7! ways. {0 cannot be at extreme left} Hence, there are 8! + 4(8! - 7!) = (36)(7!) numbers in the desired category
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