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.

_{1}×C

_{2}×…×C

_{n}

When there are including

mevents 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 =

^{4}C

_{3}×

^{6}C

_{4}=60.

Case 2: Students answer 4 out of first 4 questions. Number of ways =

^{4}C

_{4}×

^{6}C

_{3}=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

^{3}C

_{1}×

^{6}C

_{2})+(

^{3}C

_{2}×

^{6}C

_{1})+

^{3}C

_{3}

=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

^{7}C

_{3}×

^{6}C

_{2})+(

^{7}C

_{4}×

^{6}C

_{1})+

^{7}C

_{5}

=(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

^{6}C

_{1}×

^{4}C

_{3})+(

^{6}C

_{2}×

^{4}C

_{2})+(

^{6}C

_{3}×

^{4}C

_{1})+

^{6}C

_{4}

=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

**3 from each pant.**

**at least**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

^{5}C

_{3}×

^{7}C

_{5}ways.

4 questions from part I and 4 questions from part II can be selected in

^{5}C

_{4}×

^{7}C

_{4}ways.

5 questions from part I and 3 questions from part ll can be selected in

^{5}C

_{5}×

^{7}C

_{3}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).

**4 women**

**at most**Solution:

**4 women means that women are less than or equal to 4. So the number of possible combinations is**

**at most**^{10}C

_{0}∙

^{8}C

_{7}+

^{10}C

_{1}∙

^{8}C

_{6}+

^{10}C

_{2}∙

^{8}C

_{5}+

^{10}C

_{3}∙

^{8}C

_{4}+

^{10}C

_{4}∙

^{8}C

_{3}

=280+2520+8400+11760+8=22968

(iii)

**4 women**

**at least**Solution:

**4 women means that women are greater than or equal to 4.**

**at least**^{10}C

_{4}×

^{8}C

_{3}+

^{10}C

_{5}×

^{8}C

_{2}+

^{10}C

_{6}×

^{8}C

_{1}+

^{10}C

_{7}×

^{8}C

_{0}

=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

^{4}C

_{3}×

^{9}C

_{4}, ways.

4 girls and 3 boys can be selected in

^{4}C

_{4}×

^{9}C

_{3}ways.

Therefore, in this case, required number of ways

^{4}C

_{3}×

^{9}C

_{4}+

^{4}C

_{4}×

^{9}C

_{3}=504+84=588

(iii) Since

**3 girls are to be there in every committee, the committee can consist of**

**at most**(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

^{4}C

_{3}×

^{9}C

_{4}ways.

2 girls and 5 boys can be selected in

^{4}C

_{2}×

^{9}C

_{5}, ways.

1 girl and 6 boys can be selected in

^{4}C

_{1}×

^{9}C

_{6}ways.

No girl and 7 boys can be selected in

^{4}C

_{0}×

^{9}C

_{7}ways

Therefore, in this case, required number of ways

Therefore, in this case, required number of ways equals the sum of the four outcomes.

**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

^{6}C

_{1}×1!+

^{6}C

_{2}×2!+

^{6}C

_{3}×3!+

^{6}C

_{4}×4!+

^{6}C

_{5}×5!+

^{6}C

_{6}×6!=1956

This can be written as the permutations. The number of permutations of 6 letters, taken 1 to 6 is given by

^{6}P

_{1}+

^{6}P

_{2}+

^{6}P

_{3}+

^{6}P

_{4}+

^{6}P

_{5}+

^{6}P

_{6}=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 ^{7}C_{5} 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

^{7}C

_{1}×

^{4}C

_{4}ways.

2 boys and 3 girls can be selected in

^{7}C

_{2}×

^{4}C

_{3}ways.

3 boys and 2 girls can be selected in

^{7}C

_{3}×

^{4}C

_{2}ways.

4 boys and 1 girl can be selected in

^{7}C

_{4}×

^{4}C

_{1}ways.

Therefore, the required number of ways

^{7}C

_{1}∙

^{4}C

_{4}+

^{7}C

_{2}∙

^{4}C

_{3}+

^{7}C

_{3}∙

^{4}C

_{2}+

^{7}C

_{4}∙

^{4}C

_{1}

=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

^{4}C

_{3}×

^{7}C

_{2}ways.

4 girls and 1 boy can be selected in

^{4}C

_{4}×

^{7}C

_{1}ways.

Therefore, the required number of ways

^{4}C

_{3}∙

^{7}C

_{2}+

^{4}C

_{4}∙

^{7}C

_{1}=91