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1a:T1947,ジョーノさんこんにちは。
確かに試験の時の時間の使い方は学校で教わらないので分からないですよね。今回は試験のときの時間配分と数学における実験の2つについて主に話していきますね。

まず、時間配分についてです。これは数学に限らず理科や共通テストのときにも応用が効きます。その解き方とは、縦ではなく横に解くということです。何を言ってるのか分からないと思うので詳しく解説していきます。

数学や理科などの科目には大問があり、それぞれの設問が紐で結ばれてるように関係がある場合が多いです。このとき、大問１つを完全に解き切ってから次の大問に進む、これを縦に解くといいこれは試験中やってはいけません。これがなぜかと言うと最終的に実力的に解けた問題、或いは簡単な問題を時間がなくて解けなかったと言う事態を避けたいからです。試験中に大問1つにハマりすぎて他の簡単な問題解く時間がなかったという経験ありませんか？それは縦に解いているからそういう状況に陥ってしまうのです。

試験において1番大事なことは、みんなが取れるような問題を落とさないということです。それが合否を左右します。難問ではあまり差がつきません。みんなが取れるような問題を落とさないことが合格へと繋がるのです。

例えば試験時間が120分で大問が6個あるとします。１つの大問に使える時間は平均して20分。その20分を一度に全部使い切るのではなく、最低でも2周をして解くのです。1周目で解ききらなくて大丈夫です。具体的には

大問1(5分)→大問2(6分)→･･･→大問6(7分)→ここから2周目→大問2(14分)で完答→･･･→大問1(15分)で完答

のような流れです。注目してほしいのが、2周目では順番が1からではないことです。これはどういうことか。1周目で何をしているのかというと、問題1つ1つの難易度、解きやすさを実験(後述)をすることで確認しています。そして解きやすいと感じた大問から2周目は解き始めるのです。

6人の人と初めましてをするとき、1人1人に20分も使いませんよね。まずは、6人の人達と軽く挨拶をして、2周目でしっかりと話をします。それと同じことです。

理解して頂けたでしょうか。数学は縦に解くのではなく、横に解くのです。

次に「実験」についてです。数学ができる人は実験を無意識のうちにやっています。数学者が未知の問題を解くときもこの実験をして解いているのです。

実験は主に以下の3つに分かれます。①文字を具体化②イメージをする③類題を作る

①文字の具体化
これは文字通り、問題で変数(x,y,z,a,s,t…etc)が与えられたとき、実際にその変数に0や1などの簡単な数字を入れて考えてみるということです。文字が増えるとどうしても難しく感じてしまいます。だから具体的な数字を入れて、簡単な問題に帰着させるのです。そして、具体的な数字で解けたのならば、その問題は解けたも同然です。なぜなら、その数字を文字に置き換えるだけで解けたことになるからです。文字に具体的な数をいれて使った解法を”そのまま”使えばいいのです。簡単に思えてきませんか。そして、最後に「検算」をやりましょう。出てきた答えに先ほど実験で使った数をその文字に代入して、実験のときに出てきた答えと一致すればその答えの正当性は増します。最初から解き直したり考え直すことをしなくても自分の答えに自信を持てるわけです。実験をやる意味を理解して頂けたでしょうか。

②イメージをする
特にグラフの問題などではイメージをすることが大切です。どういうことか。関数の問題を数式だけで解くのではなく、グラフを使ってイメージをしながら解くということです。代数的に解くのではなく、幾何学的にまず考えてそれでも無理なら代数的に解きましょう。

共通テストの問題は解けるのに、2次試験の問題が解けないという悩みがよくありますが、それは分野ごとに解いているからです。
難関大レベルにもなると、二次関数の問題で複素数やベクトル、三角関数を使う必要が出てきます。つまり本当の「数学力」が要求されているわけですね。二次関数の問題は代数的に、ベクトルはベクトルを使って、複素数の問題では複素数を使うなど分野ごとに縛られるのではなく色々な分野に跨って考えてください。

③類題をつくる
これは実験の中で1番難しく、出来るようになるには経験が必要になります。例えば、3元1次不定方程式の問題(過去に難関大で出題された)

「35x+91y+65z=3の満たす整数(x,y,z)の組を1組求めよ。」

というもの。これは一見難しそうに見えて解法が思いつきません。しかし類題を作ることによって解き方が見えてきます。類題作れますか？今回の場合は三元→二元に変えて少し簡単な類題を作りましょう。そして

「2x+3y=5を満たす整数の組を1組求めよ。」

という類題を解いてみます。そのとき「合同式を使ってみよう！」と閃いたのならば、「お！では三元の場合も同じように合同式を使ってみよう！」とアイデアが浮かぶわけです。

解法が思い浮かぶかは運要素がありますが、実験のうちの「類題をつくる」をすることで解法を思い浮かびやすくなりましたね。

以上、説明できることの殆どを説明することができました。あとは実践あるのみです。たくさんの問題を解く上で今日学んだことを意識して“質の高い”演習を積んでください。

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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"確率Pの実用性"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"具体例を使って説明していきます。\n\nまず階乗n!についてです。\n生徒8人を1列に並べる方法は何通りか\nA. 8!通り\n階乗はその人数の並べ方がいくつかを表しています。\n\n次にnPrについてです。\n生徒が8人おり、そのうち5人を選んで並べる方法は何通りか\nA. 8P5通り(Pの左の数が8で右の数が5ということです。)\nnPrはある人数の人達から何人かを選んで並べる方法がいくつかを表しています。\n\n最後にnCrについてです。\n生徒が8人おり、そのうち5人を選ぶ方法は何通りか\nA. 8C5通り\nnCrはある人数の人達から何人かを選ぶ方法がいくつかを表しています。\n\n階乗とnPrとnCrについて説明しましたがひとつ気づくことはありませんか。\n日本語で表すと\nnPrは選んでそれらを並べる方法\nnCrは選び方\n階乗は並べ方\nです。つまりnPrにnCrと階乗が含まれています。\n式で表すと8P5＝8C5×5!\n右辺が表しているのは8人の中から5人を選び(8C5)、その人たちを並べる(5!)方法を表しています。8P5と意味は全くおなじですね。\n\n具体例で説明してきましたが、一般化すると\nnPr＝nCr×r!\nというような関係になります。実際にやれば分かりますが、式を展開しても右辺も左辺も同じ形になります。なのでPを一切使わなくてもCや階乗などを使えばどの問題も解けます。確かに言われてみればPというのはなくてもいいものですね。個人の予想ですが、Pを使うと答案を書くのが楽になるから色んな参考書の解答で書いてあるのでしょうかね。\n\nPを使うかどうかは自由ですが、PとCを使い分けるのが面倒くさいなら、問題が選び方を指してるのか並べ方を指しているのかを意識しながらCや階乗で解くのがいいでしょう。個人的にはPを使わないでとくのがおすすめです！もちろん使ったからといって悪いわけではないですよ！"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 3.31-2 6-2m0-12C9.79 4 8 5.79 8 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overflow-hidden","children":["$","div",null,{"className":"w-full h-full 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"整数、確率が苦手すぎます"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんばんは！回答が遅くなってすみません！\n確率と整数は自分自身得意としていたので、答えさせていただきます！\n\nまずいずれにも共通することですが、青チャートやFocus goldの例題レベルの問題は確実に網羅してください！この基礎が発展的な問題を解くのに必ず必要になります。ですので、まずこの部分が抜けていたら、確認しておくようにしてください！\n\n確率についてです。確率はとにかく事象を文章から図などへと可視化してみましょう！どのような動きが起こっているのか、どのような特徴があるのか、それを分かりやすくするのです。この動作は慣れないといけないので、何問かやってみるといいです。とにかく文章を分かりやすくしてみることを意識してみてください。文章では見えなかった情報が浮かび上がってくるはずです。また、それでも厳しければ、青チャートなどの例題で似た問題がないか思い浮かべ、その解き方を真似してみてください！\nそれを何問しても厳しければ、「合格る確率」という参考書を購入して、stage3のみやることをおすすめします。stage3のみといったのは、それが最も入試に頻出かつ全パターンを網羅しているいいレベルの問題だからです。そのパターンを全て覚えましょう！\n\n次に、整数です。整数は確率より厄介な問題が多いですが、基本的にはあのユークリッドの互除法を用いた不定方程式の解法パターンを覚えておけば基本はOKです。そして整数問題が出てきた時は、このパターンの発展系ではないか確認してみましょう。\nまた、整数では、学校では言われないよく出るパターンがあるのです。例えば、平方数はmod3、4を使うと数を絞れる、などです。これは、YouTubeのPASSLABOというチャンネルで整数全パターン解説という動画で整数を全部網羅してみるので、よかったらやってみてください！(回し者ではないです笑)\n\n確率と整数は個人的に数3よりも難しいなと思います。ですので、まとまった時間をとって集中的にやってみるといいです！応援しています！"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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10.05z","children":[]}]]],"style":{"color":"$undefined"},"height":16,"width":16,"xmlns":"http://www.w3.org/2000/svg"}],["$","div",null,{"className":"text-xs mr-2","children":4}]]}],["$","div",null,{"className":"text-xs rounded-lg border border-text px-4","children":"理系数学"}]]}]]}],["$","div",null,{"className":"ml-3","children":["$","div",null,{"className":"bg-caption w-20 h-20 rounded-sm overflow-hidden","children":["$","div",null,{"className":"w-full h-full 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"質問 何を数え落とした"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"かきふらいさん初めまして。\n文系ですが、回答させていただきます。\n\n本回答の構成は以下です。\n①数え落とした5通りはなにか\n②再考察が合っているか\n③数え上げる際のポイント\n\n以上3点です。\n\n①数え落とした5通りはなにか\n結論としては、端が白玉になる場合が想定されていないかなと思います。\n\nex.●◯●●◯●●◯●●◯●◯●◯\n\n上記のような場合です。\n\n②再考察が合っているか\n結論、誤りです。\n本問の場合、同色の玉に区別は存在しません。\n\n区別する場合は玉に番号が振ってあったりする場合でしょう。本問のように、玉を「色」でしか判別していない場合、区別する必要がないのです。\n\n③数え上げる際のポイント\n質問が若干抽象的なので、回答者様の解法に近いもので解く場合、私なら以下のように解きます。\n\nまずは、解答のための必要条件を具体的に導きます。\n必要条件は、\n「白玉同士は隣接しない」\n「赤玉は白玉に隣接する」\n以上の2つです。\n\nしたがって、初期配置は以下のようになります。\n\n◯●◯●◯●◯●◯●◯\n\n上記のように、赤5個、白6個を配置できます。残る赤4個をどこに配置するかですが、以下のようになります。\n\n1◯●2◯●3◯●4◯●5◯●6◯7\n\nつまり7箇所、赤を置く場所があります。なお、１つの場所につき置けるのは1つまでです。一見、「◯●」の間に置く場合も考えた方が良さそうな気がしますが、これは考慮する必要がありません。なぜなら、先ほど述べた通り、同色の玉はそれを区別する必要がないからです。\n例えば、\n「1◯●2」の2に赤を置けば、「1◯●●」となります。\n次に、\n「1◯●2」の◯●の間に●を置く場合を想定しても「1◯●●」となります。\nこのように、結果が同じになる事が分かるかと思います。\n\nしたがって、7箇所のスペースに4つの赤を置くケースを想定するので、\n\n7C4＝35\n\nよって、35通りとなります。"}],["$","div",null,{"className":"flex 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"場合の数・確率がどうしてもできません"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"苦手分野だからといって、捨てることはあまりオススメしません。\n\n残り時間が少ないから、限られた勉強時間でとれる得点を上げるために…と考えたくなるのはよくわかります。選択問題では苦手分野を回避できますが、2次試験では全員に同じ問題が課されます。たとえば、基礎レベルの確率分野の問題でるかもしれません。難易度が高くないなら正答率は高くなるでしょう。ですが、苦手分野だからといって捨てれば、周りとかなり差をつけられてしまいます。ですので、苦手分野であれど、苦手なりに対策しておくことが大切になってくると思います。\n\nたとえばセンターの選択問題で、時間を測るときは図形を選ぶが、あとで確率分野も解いてみる。青チャート等の問題を全て解くのは厳しいけど、このレベルまでは解けるようにしておく。など、苦手なりではあるけれど、対策をしてほしいと思います。少しでもやっときゃよかった…と後悔してるときには遅いので…\n\n残りの期間、頑張ってください。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"確率が分からなすぎる"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"確率の定義をとても簡単に言えば、\n（その事象が起こる場合の数）／（起こりうるすべての場合の数）\nです。つまり、これが確率の最たる基本原則です。なので、確率の問題を解くにあたっては、①まずは試行の結果としてどんな場合が起こりうるかを調べ、②次に問題で問われている特定の結果が生じる場合としてどんな場合があるかを調べ、③両者の場合の数の割合を分数で表す、という三段階の作業が《基本》になります。和事象、積事象、余事象、排反、独立試行、反復試行、条件付き確率、確率の分野は色んな公式や法則が出てきますが、いずれもこの《基本》の上に成り立つものです。ですから、どんな問題もまずは《基本》の3ステップを心がけると良いと思います（私もそうしてました）。公式や問題の解説も、まず全体が何なのか、そして問われているのはどういう結果が生じる場合なのか、この二つの視点に重きを置きながら読んでいくと、その公式や問題に対する理解も深まると思います。あとは場合の数を正確に求められるように基本レベルの問題で訓練し、それと一緒に模試や過去問、レベルの高い問題集などで発展的な問題の場数も重ねていけば、「わかる」という経験もそのうち増えていって、確率を得意分野にすることも夢じゃないと思います。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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