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×…×C_n

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

=(C_E1×…)+(C_E2×…)+⋯+(C_Em×…)

Remember this,

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

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

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

(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
¹⁰C₀∙⁸C₇+¹⁰C₁∙⁸C₆+¹⁰C₂∙⁸C₅+¹⁰C₃∙⁸C₄+¹⁰C₄∙⁸C₃
=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:

(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

(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
=⁷C₁∙⁴C₄+⁷C₂∙⁴C₃+⁷C₃∙⁴C₂+⁷C₄∙⁴C₁
=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
= ⁴C₃∙⁷C₂+⁴C₄∙⁷C₁=91