12th Grade > Mathematics
PERMUTATIONS AND COMBINATIONS MCQs
Permutations And Combinations
Total Questions : 60
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Answer: Option A. -> 185
:
A
Required number of ways = 12C3−7C3
= 220 - 35 = 185
:
A
Required number of ways = 12C3−7C3
= 220 - 35 = 185
Answer: Option A. -> 7200
:
A
The total number of 4 digits are 9999-999=9000.
The numbers of 4 digits number divisible by 5 are 90×20=1800. Hence required number of ways are 9000-1800 =7200.
{Since there are 20 numbers in each hundred (1 to 100) divisible by 5 and from 999 to 9999 there are 90 hundreds, hence the results}.
:
A
The total number of 4 digits are 9999-999=9000.
The numbers of 4 digits number divisible by 5 are 90×20=1800. Hence required number of ways are 9000-1800 =7200.
{Since there are 20 numbers in each hundred (1 to 100) divisible by 5 and from 999 to 9999 there are 90 hundreds, hence the results}.
Answer: Option D. -> 28
:
D
The number of ways selecting 6 out of 10 so that 2 particular players are always excluded is 10−2C6
:
D
The number of ways selecting 6 out of 10 so that 2 particular players are always excluded is 10−2C6
Answer: Option B. -> 20
:
B
Number of diagonals in an Octagon = 8C2−8 = 20
:
B
Number of diagonals in an Octagon = 8C2−8 = 20
Answer: Option B. -> 900
:
B
The word ARRANGE, has AA,RR, NGE letters. That is two A' s, two R's and N,G,E one each.
∴The total number of arrangements 7!2!2!1!1!1!=1260
But, the number of arrangements in which bothRR are together as one unit = 6!2!1!1!1!1! = 360
∴The number of arrangements in which both RR do not come together = 1260 -360 = 900.
:
B
The word ARRANGE, has AA,RR, NGE letters. That is two A' s, two R's and N,G,E one each.
∴The total number of arrangements 7!2!2!1!1!1!=1260
But, the number of arrangements in which bothRR are together as one unit = 6!2!1!1!1!1! = 360
∴The number of arrangements in which both RR do not come together = 1260 -360 = 900.
Answer: Option C. -> n(n−1)(n−2)(n−3)8
:
C
Since no two lines are parallel and no three are concurrent, therefore n straight lines intersect at nC2 = N(say) points. Since two points are required to determine a straight line, therefore the total number of lines obtained by joining N points NC2. But in this each old line has been counted n−1C2 times, since on each old line there will be n-1 points of intersection made by the remaining (n-1) lines.
Hence the required number of fresh lines is NC2−n.n−1C2 = N(N−1)2−n(n−1)(n−2)2
= nC2(nC2−1)2−n(n−1)(n−2)2 = n(n−1)(n−2)(n−3)8
:
C
Since no two lines are parallel and no three are concurrent, therefore n straight lines intersect at nC2 = N(say) points. Since two points are required to determine a straight line, therefore the total number of lines obtained by joining N points NC2. But in this each old line has been counted n−1C2 times, since on each old line there will be n-1 points of intersection made by the remaining (n-1) lines.
Hence the required number of fresh lines is NC2−n.n−1C2 = N(N−1)2−n(n−1)(n−2)2
= nC2(nC2−1)2−n(n−1)(n−2)2 = n(n−1)(n−2)(n−3)8
Answer: Option A. -> (2n)!n!
:
A
1.3.5......(2n−1)2n = 1.2.3.4.5.6....(2n−1)(2n)2n2.4.5.....2n
= (2n)!2n2n(1.2.3......n) = (2n)!n!
:
A
1.3.5......(2n−1)2n = 1.2.3.4.5.6....(2n−1)(2n)2n2.4.5.....2n
= (2n)!2n2n(1.2.3......n) = (2n)!n!
Answer: Option B. -> 20
:
B
Number of diagonals in an Octagon = 8C2−8 = 20
:
B
Number of diagonals in an Octagon = 8C2−8 = 20
Answer: Option D. -> 24
:
D
By fundamental theorem of Multiplication = 4 × 3 × 2
:
D
By fundamental theorem of Multiplication = 4 × 3 × 2
Answer: Option B. -> 192
:
B
The integer greater than 6000 may be of 4 digits or 5 digits. So, here two cases arise.
Case I When number is of 4 digits.
Four digit number can start from 6, 7 or 8.
Thus, total number of 4-digit numbers, which are greater than
6000=3×4×3×2=72
Case II When number is of 5 digits.
Total number of five digit numbers which are greater than 6000 = 5! = 120
∴ Total number of integers = 72 + 120 = 192
:
B
The integer greater than 6000 may be of 4 digits or 5 digits. So, here two cases arise.
Case I When number is of 4 digits.
Four digit number can start from 6, 7 or 8.
Thus, total number of 4-digit numbers, which are greater than
6000=3×4×3×2=72
Case II When number is of 5 digits.
Total number of five digit numbers which are greater than 6000 = 5! = 120
∴ Total number of integers = 72 + 120 = 192