Combinations with the phrases ‘at least’ or ‘at most’

Outcome is equal to the product of several combinations of multiple variables.
If there are n variables of an event then the outcome is equal to the product of combinations of n variables.

outcome=C1×C2×…×Cn

When there are including m events of compound events with the phrase ‘at least’ or ‘at most’ or other similar meaning then the outcome is given by

=(CE1×…)+(CE2×…)+⋯+(CEm×…)

Remember this,

Number of Combinations:

The number of all combinations of n things, taken r at a time is:
combinations

Examples and Solutions on cpmbinations with the phrase ‘at least’ or ‘at most’

Example 1:
In an examination, a student is supposed to answer 7 out of 10 questions in such a way that he needs to answer at least 3 out of first four. Then find the number of ways, in which a student can answer the paper?
A. 20 B. 60 C. 80 D. 100
correct: C, solution:
Case I: Students answer 3 out of first 4 questions. Number of ways =4C3×6C4=60.
Case 2: Students answer 4 out of first 4 questions. Number of ways =4C4×6C3=20.
Total number of ways =60+20=80 ways.

Example 2:
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
∴ Required number of ways

=(3C1×6C2 )+(3C2×6C1 )+3C3
=45+18+1=64

Example 3:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Solution:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways

=(7C3×6C2 )+(7C4×6C1 )+7C5
=(525+210+21)=756

Example 4:
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Solution:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways

=(6C1×4C3 )+(6C2×4C2 )+(6C3×4C1 )+6C4
=24+90+80+15=209

Example 5:
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part ll, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the
questions?
Solution:
It is given that the question paper consists of 12 questions divided into two parts — Part I and Part II, containing 5 and 7 questions, respectively.
A student has to attempt 8 questions, selecting at least 3 from each pant.
This can be done as follows.
(a). 3 questions from part I and 5 questions from part II
(b). 4 questions from part I and 4 questions from part II
(c). 5 questions from part I and 3 questions from part II
3 questions from part I and 5 questions from part II can be selected in 5C3×7C5 ways.
4 questions from part I and 4 questions from part II can be selected in 5C4×7C4 ways.
5 questions from part I and 3 questions from part ll can be selected in 5C5×7C3 ways.
Thus, required number of ways of selecting questions

combinations at least

Example 6:
There are 8 men and 10 women members of a club. How many committees of them seven can be formed, having:
(i). 4 women
Solution:
Since we have to form combinations of 4 women out of 10
women and 3 men out 8. Therefore
combinations at most

(ii). at most 4 women
Solution:
at most 4 women means that women are less than or equal to 4. So the number of possible combinations is
10C08C7+10C18C6+10C28C5+10C38C4+10C48C3
=280+2520+8400+11760+8=22968

(iii) at least 4 women
Solution:
at least 4 women means that women are greater than or equal to 4.
=10C4×8C3+10C5×8C2+10C6×8C1+10C7×8C0
=11760+7056+1680+120=20616

Example 7:
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done
when the committee consists of:
(i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?
Solution:
(i) A committee of 7 has to be formed from 9 boys and 4 girls.
Since exactly 3 girls are to be there in every committee, each committee must consist of
(7-3)=4 boys only.
Thus, in this case, required number of ways:

exactly combinations

(ii) Since at least 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or (b) 4 girls and 3 boys
3 girls and 4 boys can be selected in 4C3×9C4, ways.
4 girls and 3 boys can be selected in 4C4×9C3 ways.
Therefore, in this case, required number of ways
4C3×9C4+4C4×9C3=504+84=588

(iii) Since at most 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys; (b) 2 girls and 5 boys; (c) 1 girl and 6 boys; (d) No girl and T boys.
3 girls and 4 boys can be selected in 4C3×9C4 ways.
2 girls and 5 boys can be selected in 4C2×9C5, ways.
1 girl and 6 boys can be selected in 4C1×9C6 ways.
No girl and 7 boys can be selected in 4C0×9C7 ways
Therefore, in this case, required number of ways
Therefore, in this case, required number of ways equals the sum of the four outcomes.
=504+756+336+36=1632

Example 8:
Find the maximum number of words (with or without meaning) that can be formed by using any number of alphabets of the word CAMPUS.
A. 1296 B.64 C. 720 D. 1956
correct: D, solution:
That means any word may consist of at least one letter of the word CAMPUS.
Number of words will be

6C1×1!+6C2×2!+6C3×3!+6C4×4!+6C5×5!+6C6×6!=1956

This can be written as the permutations. The number of permutations of 6 letters, taken 1 to 6 is given by
6P1+6P2+6P3+6P4+6P5+6P6=1956

Example 9: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i). no girl? (ii). at least one boy and one girl? (iii). at least 3 girls?
Solution:
(i). Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in 7C5 ways. Therefore, the required number of ways

combinations of 7 things taken 5

(ii). Since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of
(a) 1 boy and 4 girls, (b) 2 boys and 3 girls
(c) 3 boys and 2 girls, (d) 4 boys and 1 girl.
1 boy and 4 girls can be selected in 7C1×4C4 ways.
2 boys and 3 girls can be selected in 7C2×4C3 ways.
3 boys and 2 girls can be selected in 7C3×4C2 ways.
4 boys and 1 girl can be selected in 7C4×4C1 ways.
Therefore, the required number of ways
=7C14C4+7C24C3+7C34C2+7C44C1
=7+84+210+140=441

(iii). Since, the team has to consist of at least 3 girls, the team can consist of
(a) 3 girls and 2 boys, or (b) 4 girls and 1 boy.
Note that the team cannot have all 5 girls, because, the group has only 4 girls.
3 girls and 2 boys can be selected in 4C3×7C2 ways.
4 girls and 1 boy can be selected in 4C4×7C1 ways.
Therefore, the required number of ways
= 4C37C2+4C47C1=91

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