Permutation and combination formula Maharashtra Board 11th
Exercise 3.1
1. A teacher want to select the class monitor in a class of 30 boys and 20 girls. In how many ways can the monitor to be selected if the monitor must be either a girl or boy?
Answer:
There are 30 boys and 20 girls a teacher can select any boy as a class monitor from 30 boys in 30 different ways so m equals to 30 a teacher can select any girl as a class monitor from 20 girls in 20 different ways so and equal to 20 therefore by fundamental principle of addition total number of ways teacher can select the class monitor equal to m + n equal to 30 + 20 = 50
Hence, there are 50 different possible ways to select a class monitor
2. A signal is generated from two flags by putting one flag above the other. If four flags of different colours are available how many different signal can be generated?
Answer:
Given 4 flags of different colour signal requires
the use of two flags the first flag can be selected from the four flags in four different page and the second flags can be selected from the remaining three flags in different ways
therefore by fundamental principle of multiplication the total number of signals generated equals to 4 into 3 equal to 12
3. How many two letter words can be formed using letters from the word 'SPACE' when repetition of letters a) allowed b) is not allowed?
Answer:.
a)
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| Permutation and combination formula exercise 3.1 MH board |
b)Numbers of letter in the word =5.
The first letter of the two letter word can be selected from the 5 letter in 4 different ways i.e., m=5.
The second letter of the two letter word can be selected from the remaining 4 letter in 4 different ways i.e., n=4. (Permutation and Combination)
Therefore, by fundamental theorem principle of multiplication the total number of two letter words equal to m×n=5 × 4=20 ways
4.How many three digit number can be formed from the digits 0, 1, 3, 5, 6 if repetition of digits
I) are allowed
2) are not allowed?
Answer :
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5. How many 3 digit number can be from formed using the digits 2, 3, 4, 5, 6 if the digits can be repeated?
Answer:
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6. A letter lock contains 3 ring and Each ring containing Pi letters. Determine maximum number of all styles that can be before the lock is open.
Answer :
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8. How many numbers between 100 and 1000 have 4 in the unit place?
Answer :
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9. How many numbers between 100 and 1000 have the digit 7 exactly once?
Answer :
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| Permutation and combination formula exercise 3.1 |
Answer :
13. A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his class room on the second floor?
Answer :
Number of ways student can go outside the school to his classroom on the second floor will be, as there are there 3 gates i.e., m=3 and 4 staircases i.e., so n=4 Hence,. by Multiplication Law= m×n=3×4=12 ways
14. How many 5 digit number formed using the digits 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Answer :
Five-digits numbers divisible by 3 are to be formed using the digits 0, 1, 2,3, 4, and 5 without repetition.
Five-digits numbers divisible by 3 are to be formed using the digits 0, 1, 2,3, 4, and 5 For a number to be divisible by 3, the sum of its digits should be divisible by 3
Consider, the digits: 1,2, 3, 4 and 5 Sum of the digits = 1+2+3+4+5=15 is divisible by 3. Any5-digit number formed using the digits 1,2,1 Starting with the most significant digit, 5 digits are available for this place. Since, repetition is not allowed, for the next significant place, 4 digits are available. Similarly, all the places can be filled as:
Number of 5-digit numbers= 5x4x3x2x1=120
Now, consider the digits:
0, 1, 2, 4 and 5 Sum of the digits 0+1+2+4+5=12 any 5 digit number formed using 0 1 2 4 5 is always divisible by 3 Starting with the most significant digit, 4 digits are available for this place (since 0 cannot be used). Since, repetition is not allowed, for the next significant place, 4 digits are available. (since 0 can now be used).
Similarly, all the places can be filled as:
Number of 5-digit numbers= 4x4x 3x 2x1 = 96
Next, consider the digits: 0, 1, 2, 3,4
Sum of the digits = 0+1 +2+3+4=10 which is not divisible by 3.
Therefore,None of the 5-digit numbers formed using the digits 0, 1,2, 3, and 4 will not be divisible by 3. Further, no other selection of 5 digits (out of given 6 digit) will give a 5-digit number, which is divisible by 3.
Total number of 5 a digit numbers divisible by 3=20+96=216
Related Posts :-
Permutation Combination Exercise 3.1
Permutation Combination Exercise 3.2
Permutation Combination Exercise 3.3
Permutation Combination Exercise 3.4
Permutation Combination Exercise 3.5
