Selecting, then arranging

Conditions on a selection

Real selections come with rules. Farhan's school picks a cricket team of $11$ from $16$ players. Some rules change the count in simple ways.

Someone must be in. If the captain must play, put him in first. Choose the other $10$ from the remaining $15$: $^{15}C_{10} = 3003$ ways.

Someone must be out. If an injured player cannot play, choose all $11$ from the other $15$: $^{15}C_{11} = 1365$ ways.

No rule. $^{16}C_{11} = 4368$. And indeed $3003 + 1365 = 4368$: every team either has the captain or does not.

Choosing from separate groups

A quiz team needs $2$ boys from $6$ and $3$ girls from $5$. Choose the boys, then choose the girls. By the multiplication principle,

The two choices are made one after the other, so the counts multiply. Do not add them.

At least one

A committee of $4$ is chosen from $5$ men and $4$ women. How many have at least one woman?

The direct way: cases. Count the committees with exactly $1$, $2$, $3$ and $4$ women, then add:

The quick way: the complement. Count all committees, then take away those with no woman at all:

"At least one" is usually quicker by the complement.

Set the committee size above the number of men. The red bar drops to $0$: every committee must then have a woman.

First select, then arrange

How many words of $2$ vowels and $3$ consonants can be made from the letters of DAUGHTER? A word here is any arrangement of letters, with or without meaning.

Step 1. The letters. DAUGHTER has $3$ vowels, A, U and E, and $5$ consonants, D, G, H, T and R. All are different.

Step 2. Select. Choose $2$ vowels in $^{3}C_{2} = 3$ ways and $3$ consonants in $^{5}C_{3} = 10$ ways.

Step 3. Arrange. Each choice gives $5$ letters, which can be arranged in $5!$ ways.

Answer: $3 \times 10 \times 5! = 3600$ words.

The pattern is general: first a combination to choose, then a factorial to arrange. A permutation alone would not stop you choosing three vowels.

Practise with new numbers

Try it in Python

The program lists every committee of $4$ and keeps those with a woman. Then it checks the complement formula and counts the DAUGHTER words.

Common mistakes

• Choosing one woman first. $4 \times {}^{8}C_{3}$ counts a committee with two women twice, once for each woman picked first.
• Adding instead of multiplying. Two groups chosen together multiply: $^{6}C_{2} \times {}^{5}C_{3}$.
• Forgetting to arrange. For words or codes, multiply the selections by the arrangements.
• Forgetting the must-have person. If the captain is in, choose only $10$ more, from $15$.

Remember these

• Someone must be in: choose the rest from the others. Someone must be out: choose all from the others.
• Separate groups: multiply the counts.
• At least one: total minus none.
• Words and codes: select with $^{n}C_{r}$, then arrange with $r!$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A team of $5$ is chosen from $10$ players, and Neha must be in it. How many teams are possible?

2. Choose $2$ of $4$ maths books and $1$ of $3$ physics books. In how many ways?

3. From $6$ boys and $2$ girls, a group of $3$ is chosen. How many groups have at least one girl?

4. How many $3$-letter words, all letters different, can be made from the letters of SPRING if each word must have the vowel I?

5. Mansi says the committees with at least one woman, from $5$ men and $4$ women, number $4 \times {}^{8}C_{3} = 224$. What went wrong?